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The following graph shows the curves:

\n

$y=0.25x^2-5$ and $y=0.5x+1$

\n

{geogebra_applet('https://www.geogebra.org/m/ZMw9xPwA')}

\n

\n

Choose the correct labels of the regions:

", "advice": "

a) $y\\le 0.5x+1 $ and $y\\ge 0.25x^2-5$

\n

This is the region below the line $y=0.5x+1$ and above the curve $y=0.25x^2-5$, i.e. it is the region labelled B.

\n

\n

\n

\n

b) $y\\le 0.5x+1 $ and $y\\le 0.25x^2-5$

\n

This is the region below both the line $y=0.5x+1$ and the curve $y=0.25x^2-5$, i.e. it is the region labelled C.

\n

\n

\n

c) $y\\ge 0.5x+1$ and $y\\le 0.25x^2-5$

\n

This is the region above the line $y=0.5x+1$ but below the curve $y=0.25x^2-5$. Thus, it is the regions D and E.

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$y\\le 0.5x+1 $ and $y\\ge 0.25x^2-5$

\n

\n

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$y\\le 0.5x+1 $ and $y\\le 0.25x^2-5$

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$y\\ge 0.5x+1$ and $y\\le 0.25x^2-5$

\n

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The following graph shows the curves:

\n

$y=0.25x^2-5$ and $y=0.5x+1$

\n

{geogebra_applet('https://www.geogebra.org/m/ZMw9xPwA')}

\n

\n

Choose the correct labels of the regions:

", "advice": "

a) $y\\le 0.5x+1 $ and $y\\ge 0.25x^2-5$

\n

This is the region below the line $y=0.5x+1$ and above the curve $y=0.25x^2-5$, i.e. it is the region labelled B.

\n

\n

\n

\n

b) $y\\le 0.5x+1 $ and $y\\le 0.25x^2-5$

\n

This is the region below both the line $y=0.5x+1$ and the curve $y=0.25x^2-5$, i.e. it is the region labelled C.

\n

\n

\n

c) $y\\ge 0.5x+1$ and $y\\le 0.25x^2-5$

\n

This is the region above the line $y=0.5x+1$ but below the curve $y=0.25x^2-5$. Thus, it is the regions D and E.

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$y\\le 0.5x+1 $ and $y\\ge 0.25x^2-5$

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\n

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$y\\le 0.5x+1 $ and $y\\le 0.25x^2-5$

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$y\\ge 0.5x+1$ and $y\\le 0.25x^2-5$

\n

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Differentiate the following functions with respect to $x$:

", "advice": "

a) $y=\\simplify{{a1}x+{a2}}$

\n

When we differentiate we “lose” one $x$, and any number on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{a1}}$.

\n

\n

b) $y=\\simplify{{b1}x+{b2}}$

\n

When we differentiate we “lose” one $x$, and any number on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{b1}}$.

\n

\n

c) $y=\\simplify{{c1}x^{c2}+{c3}}$

\n

To differentiate $\\simplify{{c1}x^{c2}}$ we multiply down in front by the power, and reduce the power by $1$. The $\\simplify[]{{c3}}$ on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{c2}*{c1}x^({c2}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[]{{c2*c1}x^{c2-1}}$.

\n

\n

d) $y=\\simplify{{d1}x^{d2}+{d3}x^{d4}+{d5}x^{d6}+{d7}x^{d8}+{d9}}$

\n

To differentiate each term with a power of $x$, we multiply down in front by the power, and reduce the power by $1$. The $\\simplify[]{{d9}}$ on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{d2}*{d1}x^({d2}-1)+{d4}*{d3}x^({d4}-1)+{d6}*{d5}x^({d6}-1)+{d8}*{d7}x^({d8}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[]{{d1*d2}x^{d2-1}+{d3*d4}x^{d4-1}+{d5*d6}x^{d6-1}+{d7*d8}x^{d8-1}}$.

\n

\n

e) $y=\\simplify{{e1}x^{e2}({e3}x^{e4}+{e5}x^{e6})+{e7}x^{e8}+{e9}}$

\n

We do not know how to differentiate this in its current form. To change it into a form which we do know how to differentiate we need to multiply out the brackets.

\n

$y=\\simplify[]{{e1}*{e3}x^({e2}+{e4})+{e1}*{e5}x^({e2}+{e6})+{e7}x^{e8}+{e9}}$

\n

$y=\\simplify[]{{e1*e3}x^{e2+e4}+{e1*e5}x^{e2+e6}+{e7}x^{e8}+{e9}}$

\n

We can now differentiate each term with a power of $x$ by multiplying down in front by the power, and reducing the power by $1$. The $\\simplify[]{{e9}}$ on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{e2+e4}*{e1*e3}x^({e2+e4}-1)+{e2+e6}*{e1*e5}x^({e2+e6}-1)+{e8}*{e7}x^({e8}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[]{{(e2+e4)*e1*e3}x^{e2+e4-1}+{(e2+e6)*e1*e5}x^{e2+e6-1}+{e7*e8}x^{e8-1}}$.

\n

\n

f) $y=\\frac{\\simplify{{f1*f2}x^{f3}+{f1*f4}x^{f5}}}{\\simplify{{f1}x^{f6}}}$

\n

We do not know how to differentiate this in its current form. However, we can simplify the fraction by taking out a factor of $\\simplify[basicplus]{{f1}x^{f6}}$ out of the denominator:

\n

$y=\\frac{\\simplify[basicplus]{{f1}x^{f6}({f2}x^{f3-f6}+{f4}x^{f5-f6})}}{\\simplify{{f1}x^{f6}}}=\\simplify[basicplus]{{f2}x^{f3-f6}+{f4}x^{f5-f6}}$.

\n

We can now differentiate each term with a power of $x$ by multiplying down in front by the power, and reducing the power by $1$.

\n

$\\frac{dy}{dx}=\\simplify[basicplus]{{f3-f6}*{f2}x^({f3-f6}-1)+{f5-f6}*{f4}x^({f5-f6}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[basicplus]{{(f3-f6)*f2}x^{f3-f6-1}+{(f5-f6)*f4}x^{f5-f6-1}}$.

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$y=\\simplify{{a1}x+{a2}}$

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$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{b1}x+{b2}}$

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$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{c1}x^{c2}+{c3}}$

\n

$\\frac{dy}{dx}=$[[0]]

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\n

$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{e1}x^{e2}({e3}x^{e4}+{e5}x^{e6})+{e7}x^{e8}+{e9}}$

\n

$\\frac{dy}{dx}=$[[0]]

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Differentiate the following functions with respect to $x$:

", "advice": "

a) $y=\\simplify{{a1}x+{a2}}$

\n

When we differentiate we “lose” one $x$, and any number on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{a1}}$.

\n

\n

b) $y=\\simplify{{b1}x+{b2}}$

\n

When we differentiate we “lose” one $x$, and any number on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{b1}}$.

\n

\n

c) $y=\\simplify{{c1}x^{c2}+{c3}}$

\n

To differentiate $\\simplify{{c1}x^{c2}}$ we multiply down in front by the power, and reduce the power by $1$. The $\\simplify[]{{c3}}$ on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{c2}*{c1}x^({c2}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[]{{c2*c1}x^{c2-1}}$.

\n

\n

d) $y=\\simplify{{d1}x^{d2}+{d3}x^{d4}+{d5}x^{d6}+{d7}x^{d8}+{d9}}$

\n

To differentiate each term with a power of $x$, we multiply down in front by the power, and reduce the power by $1$. The $\\simplify[]{{d9}}$ on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{d2}*{d1}x^({d2}-1)+{d4}*{d3}x^({d4}-1)+{d6}*{d5}x^({d6}-1)+{d8}*{d7}x^({d8}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[]{{d1*d2}x^{d2-1}+{d3*d4}x^{d4-1}+{d5*d6}x^{d6-1}+{d7*d8}x^{d8-1}}$.

\n

\n

e) $y=\\simplify{{e1}x^{e2}({e3}x^{e4}+{e5}x^{e6})+{e7}x^{e8}+{e9}}$

\n

We do not know how to differentiate this in its current form. To change it into a form which we do know how to differentiate we need to multiply out the brackets.

\n

$y=\\simplify[]{{e1}*{e3}x^({e2}+{e4})+{e1}*{e5}x^({e2}+{e6})+{e7}x^{e8}+{e9}}$

\n

$y=\\simplify[]{{e1*e3}x^{e2+e4}+{e1*e5}x^{e2+e6}+{e7}x^{e8}+{e9}}$

\n

We can now differentiate each term with a power of $x$ by multiplying down in front by the power, and reducing the power by $1$. The $\\simplify[]{{e9}}$ on its own disappears. Therefore we get

\n

$\\frac{dy}{dx}=\\simplify[]{{e2+e4}*{e1*e3}x^({e2+e4}-1)+{e2+e6}*{e1*e5}x^({e2+e6}-1)+{e8}*{e7}x^({e8}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[]{{(e2+e4)*e1*e3}x^{e2+e4-1}+{(e2+e6)*e1*e5}x^{e2+e6-1}+{e7*e8}x^{e8-1}}$.

\n

\n

f) $y=\\frac{\\simplify{{f1*f2}x^{f3}+{f1*f4}x^{f5}}}{\\simplify{{f1}x^{f6}}}$

\n

We do not know how to differentiate this in its current form. However, we can simplify the fraction by taking out a factor of $\\simplify[basicplus]{{f1}x^{f6}}$ out of the denominator:

\n

$y=\\frac{\\simplify[basicplus]{{f1}x^{f6}({f2}x^{f3-f6}+{f4}x^{f5-f6})}}{\\simplify{{f1}x^{f6}}}=\\simplify[basicplus]{{f2}x^{f3-f6}+{f4}x^{f5-f6}}$.

\n

We can now differentiate each term with a power of $x$ by multiplying down in front by the power, and reducing the power by $1$.

\n

$\\frac{dy}{dx}=\\simplify[basicplus]{{f3-f6}*{f2}x^({f3-f6}-1)+{f5-f6}*{f4}x^({f5-f6}-1)}$

\n

$\\frac{dy}{dx}=\\simplify[basicplus]{{(f3-f6)*f2}x^{f3-f6-1}+{(f5-f6)*f4}x^{f5-f6-1}}$.

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$y=\\simplify{{a1}x+{a2}}$

\n

$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{b1}x+{b2}}$

\n

$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{c1}x^{c2}+{c3}}$

\n

$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{d1}x^{d2}+{d3}x^{d4}+{d5}x^{d6}+{d7}x^{d8}+{d9}}$

\n

$\\frac{dy}{dx}=$[[0]]

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$y=\\simplify{{e1}x^{e2}({e3}x^{e4}+{e5}x^{e6})+{e7}x^{e8}+{e9}}$

\n

$\\frac{dy}{dx}=$[[0]]

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Solve the following equations involving logarithms:

", "advice": "

a) $\\simplify[]{log({an^a},{an})}=x\\implies\\simplify[]{{an^a}={an}^x}$

\n

Since $\\simplify[]{{an^a}={an}^{a}}$, then $\\simplify[]{log({an^a},{an})={a}}$.

\n

\n

\n

b) $\\simplify[]{log({bn^b},{bn})}=x\\implies\\simplify[]{{bn^b}={bn}^x}$

\n

Since $\\simplify[]{{bn^b}={bn}^{b}}$, then $\\simplify[]{log({bn^b},{bn})={b}}$.

\n

\n

c) $\\simplify[]{log(x,{cn})={c}}\\implies\\simplify[]{x={cn}^{c}={ansc}}$.

\n

\n

d) $\\simplify[]{log(1/{dn^d},{dn})}=x\\implies\\simplify[]{1/{dn^d}={dn}^x}$

\n

Since $\\simplify[]{{dn^d}={dn}^{d}}$, then $\\simplify[]{1/{dn^d}={dn}^{-d}}$, thus $\\simplify[]{log(1/{dn^d},{dn})={-d}}$.

\n

\n

e) $\\simplify[]{log(x,{en})={-ee}}\\implies\\simplify[]{x={en}^{-ee}=1/({en}^{ee})=1/{anse}}$.

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$\\simplify[]{log({an^a},{an})}=x$

\n

$x=$[[0]]

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$\\simplify[]{log({bn^b},{bn})}=x$

\n

$x=$[[0]]

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$\\simplify[]{log(x,{cn})={c}}$

\n

$x=$[[0]]

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$\\simplify[]{log(1/{dn^d},{dn})}=x$

\n

$x=$[[0]]

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$\\simplify[]{log(x,{en})={-ee}}$

\n

$x=$[[0]]

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Answer the questions about the following quadratic curves:

", "advice": "

a) $y=\\simplify{x^2+{ca+cb}x+{ca*cb}}$

\n

The $y$-intercept is where the curve cuts the $y$-axis. To find the co-ordinates of this point, we use the fact that the $y$-axis is precisely the point on a graph where $x=0$. We can then use $x=0$ in the definition of the curve to find the corresponding $y$-coordinate:

\n

$y=\\simplify[]{0^2+{ca+cb}*0+{ca*cb}={ca*cb}}$.

\n

Therefore the co-ordinates of the $y$-intercept are $(0,\\simplify[]{{ca*cb}})$.

\n

The $x$-intercepts are where the curve cuts the $x$-axis. To find the co-ordinates of this point,we use the fact that the $x$-axis is precisely the point on a graph where $y=0$. We use the definition of the curve to give

\n

$0=\\simplify{x^2+{ca+cb}x+{ca*cb}}$.

\n

This is a quadratic equation. We see we can solve it by factorising to 

\n

$0=\\simplify[]{(x+{ca})(x+{cb})}$.

\n

This has solutions

\n

$\\simplify[]{x+{ca}=0}\\implies\\simplify[]{x=-{ca}}$

\n

and

\n

$\\simplify[]{x+{cb}=0}\\implies\\simplify[]{x=-{cb}}$.

\n

The lower value of $x$ where the curve cuts the $x$-axis is therefore $\\simplify[]{-{cb}}$ and the higher value is $\\simplify[]{-{ca}}$, giving $x$-intercepts of $(\\simplify[]{{-cb}},0)$ and $(\\simplify[]{{-ca}},0)$.

\n

\n

b) $y=\\simplify{{na}*x^2+{nb}*x+{nc}}$

\n

To find the gradient we must differentiate to find $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[]{{2*na}x+{nb}}$.

\n

i) To find the gradient where $\\simplify{x={nx1}}$ we use $\\simplify{x={nx1}}$ in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[!basic]{{2*na}*{nx1}+{nb}={2*na*nx1+nb}}$,

\n

thus the gradient is $\\simplify[]{{2*na*nx1+nb}}$ at $\\simplify{x={nx1}}$.

\n

ii) To find the gradient where $\\simplify{x={nx2}}$ we use $\\simplify{x={nx2}}$ in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[!basic]{{2*na}*{nx2}+{nb}={2*na*nx2+nb}}$,

\n

thus the gradient is $\\simplify[]{{2*na*nx2+nb}}$ at $\\simplify{x={nx2}}$.

\n

\n

c) $y=\\simplify{(x+{b1})(x+{b2})}$

\n

To find the gradient we must differentiate to find $\\frac{dy}{dx}$. However this is not currently in a form we know how to differentiate. To convert it into a form we can differentiate, we multiply out the brackets to find:

\n

$y=\\simplify[]{x^2+{b1}x+{b2}x+{b1}*{b2}=x+{b1+b2}x+{b1*b2}}$. Then

\n

$\\frac{dy}{dx}=\\simplify[]{2x+{b1+b2}}$.

\n

i) To find the gradient where $\\simplify{x={bx}}$ we use $\\simplify{x={bx}}$ in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[]{2*{bx}+{b1+b2}={2*bx+b1+b2}}$,

\n

thus the gradient is $\\simplify[]{{2*bx+b1+b2}}$ at $\\simplify{x={bx}}$.

\n

ii) To find the gradient where the curve cuts the $x$-axis, we need to obtain $x$-values to input into our formula for $\\frac{dy}{dx}$. The graph cuts the $x$-axis where $y=0$. We can use our original factorised form of our function to solve:

\n

$0=\\simplify{(x+{b1})(x+{b2})}$

\n

$\\simplify[]{x+{b1}=0}$ or $\\simplify[]{x+{b2}=0}$

\n

$\\simplify[]{x=-{b1}}$ or $\\simplify[]{x=-{b2}}$.

\n

We use the lower value of $x$ first in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[]{2*{-b2}+{b1+b2}={b1-b2}}$

\n

thus the gradient is $\\simplify[]{{b1-b2}}$ at $\\simplify{x={-b2}}$, and

\n

$\\frac{dy}{dx}=\\simplify[]{2*{-b1}+{b1+b2}={b2-b1}}$

\n

thus the gradient is $\\simplify[]{{b2-b1}}$ at $\\simplify{x={-b1}}$.

\n

\n

d) Let $y=(\\simplify[fractionNumbers]{x-{a}})^2$

\n

To find the gradient we must differentiate to find $\\frac{dy}{dx}$. However this is not currently in a form we know how to differentiate. To convert it into a form we can differentiate, we multiply out the brackets to find:

\n

$y=(\\simplify[fractionNumbers]{x-{a}})(\\simplify[fractionNumbers]{x-{a}})$

\n

$y=\\simplify[fractionNumbers]{x^2-{2a}x+{a*a}}$.

\n

We then differentiate:

\n

$\\frac{dy}{dx}=\\simplify[fractionNumbers]{2x-{2*a}}$.

\n

i) To find the gradient where $\\simplify{x={ax1}}$ we use $\\simplify{x={ax1}}$ in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[!basic]{2*{ax1}-{2*a}={2*ax1-2*a}}$,

\n

thus the gradient is $\\simplify[]{{2*ax1-2*a}}$ at $\\simplify{x={ax1}}$.

\n

ii) To find the gradient where $\\simplify{x={ax2}}$ we use $\\simplify{x={ax2}}$ in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[!basic]{2*{ax2}-{2*a}={2*ax2-2*a}}$,

\n

thus the gradient is $\\simplify[]{{2*ax2-2*a}}$ at $\\simplify{x={ax2}}$.

\n

iii) To find the gradient where the curve cuts the $y$-axis, we need to obtain an $x$-value to input into our formula for $\\frac{dy}{dx}$. We use the fact that the $y$-axis is precisely the point on a graph where $x=0$. Using $x=0$ in our formula for $\\frac{dy}{dx}$ we find

\n

$\\frac{dy}{dx}=\\simplify[!basic]{2*0+{-2a}={-2a}}$,

\n

thus the gradient is $\\simplify[]{{-2a}}$ where the curve cuts the $y$-axis.

\n

\n

e) $y=\\simplify{{da}*x^2+{db}*x+{dc}}$

\n

To find the gradient we must differentiate to find $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[]{{2*da}x+{db}}$.

\n

i) To find the gradient where $\\simplify{x={dx}}$ we use $\\simplify{x={dx}}$ in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[!basic]{{2*da}*{dx}+{db}={2*da*dx+db}}$,

\n

thus the gradient is $\\simplify[]{{2*da*dx+db}}$ at $\\simplify{x={dx}}$.

\n

ii) To find the gradient where the curve cuts the $y$-axis, we need to obtain an $x$-value to input into our formula for $\\frac{dy}{dx}$. We use the fact that the $y$-axis is precisely the point on a graph where $x=0$. Using $x=0$ in our formula for $\\frac{dy}{dx}$ we find

\n

$\\frac{dy}{dx}=\\simplify[]{{2*da}*0+{db}={db}}$,

\n

thus the gradient is $\\simplify[]{{db}}$ where the curve cuts the $y$-axis.

\n

iii) To find the gradient where the curve cuts the $x$-axis, we need to obtain $x$-values to input into our formula for $\\frac{dy}{dx}$. The graph cuts the $x$-axis where $y=0$. Thus we need to solve the equation 

\n

$0=\\simplify{{da}*x^2+{db}*x+{dc}}$

\n

to find the $x$-values where $y=0$. We see we can factorise:

\n

$\\simplify[basicplus]{0=({d1}x+{d2})({d3}x+{d4})}$, giving

\n

$\\simplify[basicplus]{{d1}x+{d2}=0}$ or $\\simplify[basicplus]{{d3}x+{d4}=0}$

\n

$\\simplify[basicplus]{x=-{d2}/{d1}}$ or $\\simplify[basicplus]{x=-{d4}/{d3}}$.

\n

We use the lower value of $x$ first in our formula for $\\frac{dy}{dx}$:

\n

$\\frac{dy}{dx}=\\simplify[!basic,basicplus]{{2*da}*-{d4}/{d3}+{db}={-2*d1*d4+d1*d4+d2*d3}}$,

\n

thus the gradient is $\\simplify[basicplus]{{-2*d1*d4+d1*d4+d2*d3}}$ at $\\simplify{x=-{d4}/{d3}}$, and

\n

$\\frac{dy}{dx}=\\simplify[!basic,basicplus]{{2*da}*-{d2}/{d1}+{db}={-2*d3*d2+d1*d4+d2*d3}}$,

\n

thus the gradient is $\\simplify[basicplus]{{-2*d3*d2+d1*d4+d2*d3}}$ at $\\simplify{x=-{d2}/{d1}}$.

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Let $y=\\simplify{x^2+{ca+cb}x+{ca*cb}}$

\n

What are the co-ordinates of the $y$-intercept?

\n

([[0]],[[1]])

\n

What are the co-ordinates of the $x$-intercepts? (give the intercept with the lower $x$ co-ordinate first)

\n

([[2]],[[3]])

\n

([[4]],[[5]])

\n

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Let $y=\\simplify{{na}*x^2+{nb}*x+{nc}}$

\n

Find the gradient where

\n

i) $\\simplify{x={nx1}}$

\n

Gradient=[[0]]

\n

ii) $\\simplify{x={nx2}}$

\n

Gradient=[[1]]

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Let $y=\\simplify{(x+{b1})(x+{b2})}$

\n

Find the gradient where

\n

i) $\\simplify{x={bx}}$

\n

Gradient=[[0]]

\n

ii) Where the curve cuts the $x$-axis (input the answer corresponding to the lower value of $x$ first)

\n

Gradient=[[1]] and [[2]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{2*bx+b1+b2}", "maxValue": "{2*bx+b1+b2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{b1-b2}", "maxValue": "{b1-b2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{-b1+b2}", "maxValue": "{-b1+b2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Let $y=(\\simplify[fractionNumbers]{x-{a}})^2$

\n

Find the gradient where

\n

i) $\\simplify{x={ax1}}$

\n

Gradient=[[0]]

\n

\n

ii) $\\simplify{x={ax2}}$

\n

Gradient=[[1]]

\n

iii) Where the curve cuts the $y$-axis

\n

Gradient=[[2]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "3", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{2*ax1-2*a}", "maxValue": "{2*ax1-2*a}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{2*ax2-2*a}", "maxValue": "{2*ax2-2*a}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "-2*a", "maxValue": "-2*a", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Let $y=\\simplify{{da}*x^2+{db}*x+{dc}}$

\n

Find the gradient where

\n

i) $\\simplify{x={dx}}$

\n

Gradient=[[0]]

\n

ii) Where the curve cuts the $y$-axis

\n

Gradient=[[1]]

\n

iii) Where the curve cuts the $x$-axis (input the answer corresponding to the lower value of $x$ first)

\n

Gradient=[[2]] and [[3]]

", "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{2*da*dx+db}", "maxValue": "{2*da*dx+db}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{db}", "maxValue": "{db}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{-2*d1*d4+d1*d4+d2*d3}", "maxValue": "{-2*d1*d4+d1*d4+d2*d3}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{-2*d3*d2+d1*d4+d2*d3}", "maxValue": "{-2*d3*d2+d1*d4+d2*d3}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}]}], "allowPrinting": true, "navigation": {"allowregen": true, "reverse": true, "browse": true, "allowsteps": true, "showfrontpage": true, "showresultspage": "oncompletion", "navigatemode": "sequence", "onleave": {"action": "none", "message": ""}, "preventleave": true, "startpassword": ""}, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "feedback": {"showactualmark": true, "showtotalmark": true, "showanswerstate": true, "allowrevealanswer": true, "advicethreshold": 0, "intro": "", "reviewshowscore": true, "reviewshowfeedback": true, "reviewshowexpectedanswer": true, "reviewshowadvice": true, "feedbackmessages": []}, "contributors": [{"name": "Christopher Tedd", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1880/"}, {"name": "Radu Dragomir Manac", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2821/"}], "extensions": ["geogebra"], "custom_part_types": [], "resources": []}