# The given are Two x-intercepts y-intercept math (0,−4) Maximum at (2,4) How to create quadratic equation given y intercept, and maximum and B=8?

The given are
Two $x$-intercepts
$y$-intercept math $\left(0,-4\right)$
Maximum at $\left(2,4\right)$
How to create quadratic equation given $y$ intercept, and maximum and $B=8$?
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embraci4i
You can start by writing out the general form of a quadratic polynomial: $P\left(x\right)=a{x}^{2}+bx+c$. Saying that the y-intercept is $\left(0,-4\right)$ is the same as saying that $P\left(0\right)=-4$, which gives us $c=-4$. Now the derivative of the polynomial will be ${P}^{\prime }\left(x\right)=2ax+b$. Since the derivative of a differentiable function at a maximum is $0$, you must have ${P}^{\prime }\left(2\right)=0$ or $4a=-b$. Since the point $\left(2,4\right)$ is assumed to be in the graph of the function, we know that $P\left(2\right)=4$ or $4a+2b-4=4$, which means that $4a=8-2b$. Thus we have
Which gives $b=8$ and $a=-2$, so we get $P\left(x\right)=-2{x}^{2}+8x-4$
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deiluefniwf
Another way to do it, would be to use the fact that every quadratic equation is of the form $P\left(x\right)=\alpha \left(x-\beta {\right)}^{2}+\gamma$ for some $\alpha ,\phantom{\rule{thickmathspace}{0ex}}\beta ,\phantom{\rule{thickmathspace}{0ex}}\gamma$. When the equation is written in this form it becomes clear that $\alpha$ must be negative (otherwise the function won't have a maximum), and then the maximum is attained when $x-\beta =0$. Since the maximum is at $x=2$, we have $\beta =2$. Also, the value of $P$ at the maximum will be $4=P\left(2\right)=\alpha \left(0{\right)}^{2}+\gamma =\gamma$, so $\gamma =4$. Lastly, we know that the y-intercept is $\left(0,-4\right)$, so that $-4=P\left(0\right)=\alpha \left(0-\beta {\right)}^{2}+\gamma =4\alpha +4$, which gives $\alpha =-2$. Now we can expand the original expression and we get $P\left(x\right)=\alpha \left(x-\beta {\right)}^{2}+\gamma =-2\left(x-2{\right)}^{2}+4=-2{x}^{2}+8x-4$, so we get the same result as before.