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(You could also use this section to provide any formulae the students might want to reference.)

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(Or it could be the main, referenced part of a multi-part question.)

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Constant term in part 3.

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One root of the polynomial in part 3.

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Coefficient of x in part 3.

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This is the value for x which is substituted in part 1.

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One root of the polynomial in part 3.

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Let's look at functions of the form $f(x) = \\simplify{a x + b}$.

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Suppose $f(x) = \\simplify{{a}x+{b}}$.

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Find $f(\\var{x0})$.

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{name}'s favourite quadratic is $g(x) = \\simplify[all,!noleadingminus]{{a2}x^2 + {b2}x + {c2}}$.

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Find $g'(x)$.

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Consider the quadratic $h(x) = \\simplify{x^2 + {b3}x + {c3}}$.

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What is the set of roots of $h(x)$? Use the Numbas syntax set(a,b,c) in your answer.

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If the roots of $h(x)$ are $r$ and $s$, then $h(x) = (x-r)(x-s) = x^2 -(r+s)x +rs$.

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What does this tell us about $r$ and $s$?

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When evaluating a function $f(x)$ at a point $x=\\var{x0}$, simply substitute $x$ in the expression with the value $\\var{x0}$.

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When differentiating a polynomial with respect to $x$, take each term $a_n x^n$ and multiply its coefficient by $n$ before reducing its power by 1, giving $n a_n x^{n-1}$.

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To find the roots of a quadratic, you might be able to do it by sight by noting that the sum of the two roots should be the negative of the coefficient of $x$, while the product of the two roots should be the constant term. This is explained in more detail in the expanded part above.

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Otherwise, you can always find the root(s) of a general quadratic $a x^2 + b x + c = 0$ by using the quadratic formula:

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\$r = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}\$

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