Find the Vertex f(x)=2x^2-7x-4

$f (x) = 2 x^{2} - 7 x - 4$

Step 1

Rewrite the equation in vertex form.

Complete the square for $2 x^{2} - 7 x - 4$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 2$

$b = - 7$

$c = - 4$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 7}{2 \cdot 2}$

Simplify the right side.

Multiply $2$ by $2$ .

$d = \frac{- 7}{4}$

Move the negative in front of the fraction.

$d = - \frac{7}{4}$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = - 4 - \frac{{(- 7)}^{2}}{4 \cdot 2}$

Simplify the right side.

Simplify each term.

Raise $- 7$ to the power of $2$ .

$e = - 4 - \frac{49}{4 \cdot 2}$

Multiply $4$ by $2$ .

$e = - 4 - \frac{49}{8}$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$e = - 4 \cdot \frac{8}{8} - \frac{49}{8}$

Combine $- 4$ and $\frac{8}{8}$ .

$e = \frac{- 4 \cdot 8}{8} - \frac{49}{8}$

Combine the numerators over the common denominator.

$e = \frac{- 4 \cdot 8 - 49}{8}$

Simplify the numerator.

Multiply $- 4$ by $8$ .

$e = \frac{- 32 - 49}{8}$

Subtract $49$ from $- 32$ .

$e = \frac{- 81}{8}$

Move the negative in front of the fraction.

$e = - \frac{81}{8}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $2 {(x - \frac{7}{4})}^{2} - \frac{81}{8}$ .

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

Set $y$ equal to the new right side.

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

Step 2

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 2$

$h = \frac{7}{4}$

$k = - \frac{81}{8}$

Step 3

Find the vertex $(h, k)$ .

$(\frac{7}{4}, - \frac{81}{8})$

Step 4

Find the Vertex f(x)=2x^2-7x-4

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