Find the Vertex f(x)=6x^2+13x-5

$f (x) = 6 x^{2} + 13 x - 5$

Step 1

Rewrite the equation in vertex form.

Complete the square for $6 x^{2} + 13 x - 5$ .

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

$a = 6$

$b = 13$

$c = - 5$

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

$d = \frac{13}{2 \cdot 6}$

Multiply $2$ by $6$ .

$d = \frac{13}{12}$

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

$e = - 5 - \frac{13^{2}}{4 \cdot 6}$

Simplify the right side.

Simplify each term.

Raise $13$ to the power of $2$ .

$e = - 5 - \frac{169}{4 \cdot 6}$

Multiply $4$ by $6$ .

$e = - 5 - \frac{169}{24}$

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{24}{24}$ .

$e = - 5 \cdot \frac{24}{24} - \frac{169}{24}$

Combine $- 5$ and $\frac{24}{24}$ .

$e = \frac{- 5 \cdot 24}{24} - \frac{169}{24}$

Combine the numerators over the common denominator.

$e = \frac{- 5 \cdot 24 - 169}{24}$

Simplify the numerator.

Multiply $- 5$ by $24$ .

$e = \frac{- 120 - 169}{24}$

Subtract $169$ from $- 120$ .

$e = \frac{- 289}{24}$

Move the negative in front of the fraction.

$e = - \frac{289}{24}$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $6 {(x + \frac{13}{12})}^{2} - \frac{289}{24}$ .

$6 {(x + \frac{13}{12})}^{2} - \frac{289}{24}$

Set $y$ equal to the new right side.

$y = 6 {(x + \frac{13}{12})}^{2} - \frac{289}{24}$

Step 2

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

$a = 6$

$h = - \frac{13}{12}$

$k = - \frac{289}{24}$

Step 3

Find the vertex $(h, k)$ .

$(- \frac{13}{12}, - \frac{289}{24})$

Step 4

Find the Vertex f(x)=6x^2+13x-5

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