Let b and c be constants such that the quadratic

Let b and c be constants such that the quadratic -2x^2 + bx + c has roots and 3 + sqrt(13] and 3 - sqrt(13). Find the vertex of the graph of the equation y = -2x^2 + bx + c.

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\(y=0=(x+3+\sqrt{13})(x-3+\sqrt{13})\\ y=x^2-3x+x\sqrt{13}+3x-9+3\sqrt{13}+x\sqrt{13}-3\sqrt{13}+13\\ y=x^2+x(-3+\sqrt{13}+3+\sqrt{13})+\sqrt{13}(3-3)-9+13\\ \color{blue}y=x^2+2\sqrt{13}x+4\)

\(V(-\frac{b}{2},-(\frac{b}{2})^2+c)\\ V(-\frac{2\sqrt{13}}{2},-(\frac{2\sqrt{13}}{2})^2+4)\\ \color{blue}V(-\sqrt{13},-9)\)

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Let b and c be constants such that the quadratic -2x^2 + bx + c

has roots and 3 + sqrt(13) and 3 - sqrt(13).

Find the vertex of the graph of the equation y = -2x^2 + bx + c.

This is a concave down parabola

The axis of symmetry is obviously x=3 because this is a halfway between the two roots.

\(-2x^2 + bx + c=0\\ \text{The roots of any quadraic are given by}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\ so\\ x=\frac{-b}{-4}\pm\frac{\sqrt{b^2+8c}}{-4}\\ x=\frac{b}{4}\pm\frac{\sqrt{b^2+8c}}{-4}\\ \frac{b}{4}=3\qquad so \qquad b=12\\ x=3\pm\frac{\sqrt{144+8c}}{-4}\\ \frac{\sqrt{144+8c}}{-4}=\sqrt{13}\\ \frac{144+8c}{16}=13\\ \frac{144+8c}{16}=13\\ 8c=64\\ c=8\\~\\ \text{So the quadratic is }-2x^2 + 12x + 8\\ f(x)=-2x^2 + 12x + 8\\ f(3)=26 \)

Here is the graph that I used as a check

LaTex

-2x^2 + bx + c=0\\
\text{The roots of any quadraic are given by}\\
x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\\
so\\
x=\frac{-b}{-4}\pm\frac{\sqrt{b^2+8c}}{-4}\\
x=\frac{b}{4}\pm\frac{\sqrt{b^2+8c}}{-4}\\
\frac{b}{4}=3\qquad so \qquad b=12\\
x=3\pm\frac{\sqrt{144+8c}}{-4}\\
\frac{\sqrt{144+8c}}{-4}=\sqrt{13}\\
\frac{144+8c}{16}=13\\
\frac{144+8c}{16}=13\\
8c=64\\
c=8\\~\\
\text{So the quadratic is   }-2x^2 + 12x + 8\\
f(x)=-2x^2 + 12x + 8\\
f(3)=26