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straight line

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parabola/quadratic

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cubic

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hyperbola

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circle

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quartic

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This equation, or its graph, can be described as a

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An equation of the form $y=ax^2+bx+c$ is a quadratic and the graph of it is called a parabola.

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opens upwards (is concave up)

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opens downwards (is concave down)

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The number in front of $x^2$ is called the coefficient of $x^2$ or the leading coefficient because it belongs to the term with the highest power.

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If this coefficient is postive then the parabola will be concave up, or 'open upwards', or in other words, look like a smiley face!

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If this coefficient is negative then the parabola will be concave down, or 'open downwards', or in other words, look like a sad face!

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The graph of this equation 

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The $y$-intercept is the value of $y$ when $x=0$, that is, the value of $y$ where the graph hits the $y$-axis. To find it, substitute $x=0$ into our equation:

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\\[y=\\simplify[unitFactor,basic,fractionNumbers]{{a}0^2+{b}0+{c}}=\\var{c}.\\]

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The $y$-intercept of the graph is $y=$[[0]].

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The axis of symmetry of a parabola is the line along which you could reflect the parabola and it would look exactly the same. For the quadratic $y=ax^2+bx+c$, the equation of the axis of symmetry is \\[x=-\\dfrac{b}{2a}.\\] Using our given equation $y=\\simplify[all,fractionNumbers]{{a}x^2+{b}x+{c}}$, we get $x=\\var{axis_x}$.

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The vertex, or the turning point, or the stationary point of the parabola is on this line, so it's $x$ coordinate must be $\\var{axis_x}$ and its $y$ coordinate must be the $y$ value that corresponds to this $x$ value after susbstituting it into the equation $y=\\simplify[all,fractionNumbers]{{a}x^2+{b}x+{c}}$. That is \\[y=\\simplify[unitFactor,basic,fractionNumbers]{{a}({axis_x})^2+{b}({axis_x})+{c}}=\\var{axis_y}.\\]

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The axis of symmetry of the graph of this function is $x=$[[0]]. 

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The vertex of the graph of this function is at $\\large($ [[0]], [[1]] $\\large)$.

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For the quadratic $y=ax^2+bx+c$, the $x$-intercepts (if they exist) are given by \\[x=\\dfrac{-b}{2a}\\pm\\dfrac{\\sqrt{b^2-4ac}}{2a}.\\]

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Notice how the first term gives the axis of symmetry and the second term says how far away from the axis of symmetry to go in both directions!

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For the equation $y=\\simplify[all,fractionNumbers]{{a}x^2+{b}x+{c}}$, we have $b^2-4ac=\\simplify[all,fractionNumbers]{{b}^2-4{a}{c}}=\\var{disc}$ and so there are no $x$-intercepts. is one $x$-intercept: are two $x$-intercepts: 

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$x$$=$$\\dfrac{-b}{2a}\\pm\\dfrac{\\sqrt{b^2-4ac}}{2a}$
 
$=$$\\simplify[basic,unitFactor,fractionNumbers]{{axis_x}}\\pm\\simplify[basic,unitFactor,fractionNumbers]{sqrt{{disc}}/({2*a})}$
 
$=$$\\var{axis_x}$  $\\var{xint0}, \\, \\var{xint1}$
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The set of $x$-intercepts of the graph would be [[0]].

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Note: If there are no intercepts, enter set()

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If there is only one intercept, say $x=5$, enter set(5)

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If there are two intercepts, say $x=-2$ and $x=1.5$, enter set(-2,1.5)

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If there are three intercepts, say $x=-2$, $x=1.5$ and $x=5$, enter set(-2,1.5,5)

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A degree $n$ polynomial has at most $n-1$ bends in its graph.

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Given the degree of a polynomial is $2$, the maximum number of possible 'bends' or 'turns' in the graph is [[0]].

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You are given the equation $y=\\simplify[all,fractionNumbers]{{a}x^2+{b}x+{c}}$. 

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