Find the Vertex f(x)=4x^2+6x-1

$f (x) = 4 x^{2} + 6 x - 1$

Step 1

Rewrite the equation in vertex form.

Complete the square for $4 x^{2} + 6 x - 1$ .

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

$a = 4$

$b = 6$

$c = - 1$

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 $d = \frac{b}{2 a}$ .

$d = \frac{6}{2 \cdot 4}$

Cancel the common factor of $6$ and $2$ .

Factor $2$ out of $6$ .

$d = \frac{2 \cdot 3}{2 \cdot 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 4$ .

$d = \frac{2 \cdot 3}{2 (4)}$

Cancel the common factor.

$d = \frac{2 \cdot 3}{2 \cdot 4}$

Rewrite the expression.

$d = \frac{3}{4}$

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 $e = c - \frac{b^{2}}{4 a}$ .

$e = - 1 - \frac{6^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

Raise $6$ to the power of $2$ .

$e = - 1 - \frac{36}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = - 1 - \frac{36}{16}$

Cancel the common factor of $36$ and $16$ .

Factor $4$ out of $36$ .

$e = - 1 - \frac{4 (9)}{16}$

Cancel the common factors.

Factor $4$ out of $16$ .

$e = - 1 - \frac{4 \cdot 9}{4 \cdot 4}$

Cancel the common factor.

$e = - 1 - \frac{4 \cdot 9}{4 \cdot 4}$

Rewrite the expression.

$e = - 1 - \frac{9}{4}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$e = - 1 \cdot \frac{4}{4} - \frac{9}{4}$

Combine $- 1$ and $\frac{4}{4}$ .

$e = \frac{- 1 \cdot 4}{4} - \frac{9}{4}$

Combine the numerators over the common denominator.

$e = \frac{- 1 \cdot 4 - 9}{4}$

Simplify the numerator.

Multiply $- 1$ by $4$ .

$e = \frac{- 4 - 9}{4}$

Subtract $9$ from $- 4$ .

$e = \frac{- 13}{4}$

Move the negative in front of the fraction.

$e = - \frac{13}{4}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x + \frac{3}{4})}^{2} - \frac{13}{4}$ .

$4 {(x + \frac{3}{4})}^{2} - \frac{13}{4}$

Set $y$ equal to the new right side.

$y = 4 {(x + \frac{3}{4})}^{2} - \frac{13}{4}$

Step 2

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

$a = 4$

$h = - \frac{3}{4}$

$k = - \frac{13}{4}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{3}{4}, - \frac{13}{4})$

Step 4

Find the Vertex f(x)=4x^2+6x-1

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