How do you write $f (x) = - 2 x^{2} + 20 x - 49$ $f(x)= -2x^2+20x-49$ in vertex form?

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Answer 1 · 2017-02-28T00:23:14

$y = - 2 {(x - 5)}^{2} + 1$ $y=-2(x-5)^2+1$

Standard Form equation: $y = a x^{2} + b x + c$ $y=ax^2+bx+c$
Vertex Form equation: $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$

Where $a$ $a$ is equal to the $a$ $a$ value of the standard form equation and $(h, k)$ $(h,k)$ is equal to the vertex of the equation.

In order to convert it, let's first fill in what we know. The $a$ $a$ value of the given equation is -2.

So, we have:

$y =$ $y=$ -2 $(x - h^{2})$ $(x-h^2)$ + $k$ $k$

In order to find the vertex, you must use the equation $- \frac{b}{2 a}$ $-b/(2a)$

Looking at the standard form equation, $b = 20$ $b=20$ and $a = - 2$ $a=-2$
So plugging in, you get $- \frac{20}{2 \cdot - 2}$ $-20/(2*-2)$

Once solved, you're x value of your vertex is 5. Now, you plug 5 into your original standard form equation.

$y = - 2 {(5)}^{2} + 20 (5) - 49$ $y=-2(5)^2+20(5)-49$

Finally, your vertex is:

$(5, 1)$ $(5, 1)$

Finally, plug it into your equation.
$y = - 2 {(x - 5)}^{2} + 1$ $y=-2(x-5)^2+1$

How do you write f(x)=−2x2+20x−49f(x)= -2x^2+20x-49 in vertex form?