How do you write $f (x) = 2 x^{2} - 7 x - 4$ $f(x)=2x^2-7x-4$ in vertex form?

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How do you write $f (x) = 2 x^{2} - 7 x - 4$ $f(x)=2x^2-7x-4$ in vertex form?

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Answer 1 · 2016-08-14T04:07:01

$y = 2 {(x - \frac{7}{4})}^{2} - \frac{81}{8}$ $y=2(x-7/4)^2-81/8$

Explanation:

Given -

$y = 2 x^{2} - 7 x - 4$ $y=2x^2-7x-4$

Find the vertex $(x, y)$ $(x, y)$

$x = \frac{- b}{2 a} = \frac{- (- 7)}{2 \times 2} = \frac{7}{4}$ $x=(-b)/(2a)=(-(-7))/(2xx2)=7/4$

At $x = \frac{7}{4}$ $x=7/4$

$y = 2 {(\frac{7}{4})}^{2} - 7 (\frac{7}{4}) - 4 = - \frac{81}{8}$ $y=2(7/4)^2-7(7/4)-4=-81/8$

$x, y$ $x, y$ coordinates of the vertex are $(\frac{7}{4}, - \frac{81}{4})$ $(7/4, -81/4)$

The vertex form of the quadratic equation is -

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

We need the values of $a, h, k$ $a, h, k$

$a = 2$ $a =2$ [coefficient of $x^{2}$ $x^2$ ]
$h = \frac{7}{4}$ $h=7/4$ [x-coordinate of the vertex]
$k = \frac{- 81}{8}$ $k=(-81)/8$ [ y-coordinate of the vertex]

Now plug in the values in the equation

$y = 2 {(x - \frac{7}{4})}^{2} - \frac{81}{8}$ $y=2(x-7/4)^2-81/8$

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How do you write $f (x) = 2 x^{2} - 7 x - 4$ $f(x)=2x^2-7x-4$ in vertex form?

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Explanation:

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How do you write f(x)=2x2−7x−4f(x)=2x^2-7x-4 in vertex form?

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How do you write $f (x) = 2 x^{2} - 7 x - 4$ $f(x)=2x^2-7x-4$ in vertex form?