Find the Vertex Form f(t)=t^2+4t-14

$f (t) = t^{2} + 4 t - 14$

Step 1

Write $f (t) = t^{2} + 4 t - 14$ as an equation.

$y = t^{2} + 4 t - 14$

Step 2

Complete the square for $t^{2} + 4 t - 14$ .

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

$a = 1$

$b = 4$

$c = - 14$

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

$d = \frac{4}{2 \cdot 1}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

$d = \frac{2 \cdot 2}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot 2}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot 2}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{2}{1}$

Divide $2$ by $1$ .

$d = 2$

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

$e = - 14 - \frac{4^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

$e = - 14 - \frac{4 \cdot 4}{4 \cdot 1}$

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

$e = - 14 - \frac{4 \cdot 4}{4 (1)}$

Cancel the common factor.

$e = - 14 - \frac{4 \cdot 4}{4 \cdot 1}$

Rewrite the expression.

$e = - 14 - \frac{4}{1}$

Divide $4$ by $1$ .

$e = - 14 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = - 14 - 4$

Subtract $4$ from $- 14$ .

$e = - 18$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(t + 2)}^{2} - 18$ .

${(t + 2)}^{2} - 18$

Step 3

Set $y$ equal to the new right side.

$y = {(t + 2)}^{2} - 18$

Find the Vertex Form f(t)=t^2+4t-14

Related Questions

Support LCMGCF.com by disabling your adblocker.