Solution assignment 04 Quadratic functions and graphs

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Assignment 4

The equation of a parabola is:

$y=(x-p)(x+2p)$

For which value of $p$ does the top of this parabola lie on the line:

$y=-2x$

Solution

The coordinates of the intersection points of this parabola with the $X$ -axis are $(p,0)$ and $(-2p,0)$ . The symmetry axis lies just in the middle between these two points and thus we have:

$x_{top}=-\displaystyle\frac{1}{2}p$

The $y$ -coordinate of the top is:

$y_{top}=(-\displaystyle\frac{1}{2}p-p)(-\displaystyle\frac{1}{2}p+2p)=-\displaystyle\frac{9}{4}p^2$

The top lies on the line $y=-2x$ so we have:

$-\displaystyle\frac{9}{4}p^2=-2.-\displaystyle\frac{1}{2}p$

$\displaystyle\frac{9}{4}p^2+p=0$

$p(\displaystyle\frac{9}{4}p+1)=0$

and thus:

$p=0$ or $p=-\displaystyle\frac{4}{9}$

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Solution assignment 04 Quadratic functions and graphs

Assignment 4

Solution

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