Find the Standard Form of the Parabola f(x)=-x^2+4x-10

$f (x) = - x^{2} + 4 x - 10$

Step 1

Write $f (x) = - x^{2} + 4 x - 10$ as an equation.

$y = - x^{2} + 4 x - 10$

Step 2

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- x^{2} + 4 x - 10 = y$

Step 3

Add $10$ to both sides of the equation.

$- x^{2} + 4 x = y + 10$

Step 4

Complete the square.

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

$a = - 1$

$b = 4$

$c = 0 = y + 10$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e = y + 10$

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

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

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

Simplify the right side.

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

Factor $2$ out of $4$ .

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

Move the negative one from the denominator of $\frac{2}{- 1}$ .

$d = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$d = - 2$

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

$e = 0 - \frac{4^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

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

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

$e = 0 - \frac{4 \cdot 4}{4 \cdot - 1}$

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$e = 0 - (- 1 \cdot 4)$

Multiply $- (- 1 \cdot 4)$ .

Multiply $- 1$ by $4$ .

$e = 0 - - 4$

Multiply $- 1$ by $- 4$ .

$e = 0 + 4$

Add $0$ and $4$ .

$e = 4$

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

$- {(x - 2)}^{2} + 4 = y + 10$

Step 5

Move all terms not containing $x$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$- {(x - 2)}^{2} = y + 10 - 4$

Subtract $4$ from $10$ .

$- {(x - 2)}^{2} = y + 6$

Step 6

Divide each term in $- {(x - 2)}^{2} = y + 6$ by $- 1$ and simplify.

Divide each term in $- {(x - 2)}^{2} = y + 6$ by $- 1$ .

$\frac{- {(x - 2)}^{2}}{- 1} = \frac{y}{- 1} + \frac{6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 2)}^{2}}{1} = \frac{y}{- 1} + \frac{6}{- 1}$

Divide ${(x - 2)}^{2}$ by $1$ .

${(x - 2)}^{2} = \frac{y}{- 1} + \frac{6}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{y}{- 1}$ .

${(x - 2)}^{2} = - 1 \cdot y + \frac{6}{- 1}$

Rewrite $- 1 \cdot y$ as $- y$ .

${(x - 2)}^{2} = - y + \frac{6}{- 1}$

Divide $6$ by $- 1$ .

${(x - 2)}^{2} = - y - 6$

Step 7

Factor.

Factor $- 1$ out of $- y$ .

${(x - 2)}^{2} = - (y) - 6$

Rewrite $- 6$ as $- 1 (6)$ .

${(x - 2)}^{2} = - (y) - 1 (6)$

Factor $- 1$ out of $- (y) - 1 (6)$ .

${(x - 2)}^{2} = - (y + 6)$

Find the Standard Form of the Parabola f(x)=-x^2+4x-10

Related Questions

Support LCMGCF.com by disabling your adblocker.