Find the Vertex f(x)=x^2-10x+25

$f (x) = x^{2} - 10 x + 25$

Step 1

Rewrite the equation in vertex form.

Complete the square for $x^{2} - 10 x + 25$ .

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

$a = 1$

$b = - 10$

$c = 25$

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

$d = \frac{- 10}{2 \cdot 1}$

Cancel the common factor of $- 10$ and $2$ .

Factor $2$ out of $- 10$ .

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 5}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 5}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 5}{1}$

Divide $- 5$ by $1$ .

$d = - 5$

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

$e = 25 - \frac{{(- 10)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $- 10$ to the power of $2$ .

$e = 25 - \frac{100}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 25 - \frac{100}{4}$

Divide $100$ by $4$ .

$e = 25 - 1 \cdot 25$

Multiply $- 1$ by $25$ .

$e = 25 - 25$

Subtract $25$ from $25$ .

$e = 0$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 5)}^{2} + 0$ .

${(x - 5)}^{2} + 0$

Set $y$ equal to the new right side.

$y = {(x - 5)}^{2} + 0$

Step 2

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

$a = 1$

$h = 5$

$k = 0$

Step 3

Find the vertex $(h, k)$ .

$(5, 0)$

Step 4

Find the Vertex f(x)=x^2-10x+25

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