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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: TMJ on June 06, 2013, 10:44:22 pm

Title: year 11 extended response past exam question help?
Post by: TMJ on June 06, 2013, 10:44:22 pm: This question has just confused me so help would be excellent.

A quadratic function is defined by f: [-5,2] ----> R and the rule f(x)=ax^2+bx+c. The turning point has coordinates (-2,9). The minimum value of the function is -7. Find the values of a, b, and c.
Title: Re: year 11 extended response past exam question help?
Post by: b^3 on June 06, 2013, 10:54:56 pm: If you write it in turning point form, that is in the form $y=a(x+h)^{2}+k$ , you can find the values of $h$ and $k$ from the turning point given. Then you can do a little sketch of the graph. Since we are told the minimum is -7 and the y value of the turning point is 9, we know that our turning point cannot be a minimum, and so has to be a maximum. That is we have an inverted parabola. So you can do the little sketch, but notice that the domain for $-5\leq x\leq 2$ . Now we can notice that the end of the domain that is furthest away from out turning point is $x=2$ , so at $x=2$ the $y$ value will be our minimum, $-7$ . So now you have a point and one unknown, with the equation $y=a(x+2)^{2}+9$ , so you can solve for $a$ . What you will end up with is the equation in turning point form, so you need to expand it out, then you will get the values of $a$ , $b$ and $c$ .

Hope that helps :)

EDIT: Fixed the error.

Have a try before you check this :)
$\begin{alignedat}{1}y & =a\left(x+h\right)^{2}+k \\ \text{Turning point at }\left(-2,9\right) \\ \implies h=2,\: k=9 \\ y & =a\left(x+2\right)^{2}+9 \\ \left(2,-7\right) \\ -7 & =a\left(2+2\right)^{2}+9 \\ -16 & =16a \\ \implies a & =-1 \\ y & =-\left(x+2\right)^{2}+9 \\ & =9-\left(x^{2}+4x+4\right) \\ & =-x^{2}-4x+5 \\ \therefore a=-1,\: b=-4,\: c=5 \end{alignedat} $