# How do you find the vertex and the intercepts for y=x^2 +10x+22?

Jun 7, 2016

vertex is (-5,-3). It intersects with x-axis in two points (-5, -+sqrt3) & y-intercept 22.

#### Explanation:

We have, by completing square, $y = {x}^{2} + 10 x + 25 - 3$, i.e., $y + 3 = {\left(x + 5\right)}^{2}$. Hence vertex (-5,-3), etc.

Jun 7, 2016

Detailed solution using completing the square

$\textcolor{b l u e}{\text{Vertex } \to \left(x , y\right) \to \left(- 5 , - 3\right)}$

$\textcolor{b l u e}{\text{Exact values "x_("intercepts}} = - 5 \pm \sqrt{3}$

Approx values$\text{ "x=-6.73" and } x = - 3.27$ to 2 decimal places

#### Explanation:

Given:$\text{ } y = {x}^{2} + 10 x + 22$ ............................Equation (1)
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Write as$\text{ } y = \left({x}^{2} + 10 x\right) + 22 + k$

where $k$ is a correction that neutralises introduced error by changing the equation form. $k$ not yet known.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now we start to change thing so we need $k$. We find its value after all the changes have been made.

$\textcolor{b r o w n}{\text{Move the power from "x^2" to outside the brackets}}$

$y = {\left(x + 10 x\right)}^{2} + 22 + k$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Remove the "x" from } 10 x}$

$y = {\left(x + 10\right)}^{2} + 22 + k$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Multiply the 10 by } \frac{1}{2}}$

$y = {\left(x + 5\right)}^{2} + 22 + k$ ...............................Equation (2)
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$\textcolor{b r o w n}{\text{Determine the value of } k}$

The error comes from the 5 inside the bracket when that bracket is squared.

So ${5}^{2} + k = 0 \text{ "->" } k = - 25$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{The completed square equation}}$

So equation (2) becomes

$y = {\left(x + 5\right)}^{2} - 3$.............................Equation (3)
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$\textcolor{b r o w n}{\text{Reading directly from Equation (3) "larr "vertex}}$

${x}_{\text{vertex}} = \left(- 1\right) \times 5 = - 5$

${y}_{\text{vertex}} = - 3$

$\textcolor{b l u e}{\text{Vertex } \to \left(x , y\right) \to \left(- 5 , - 3\right)}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(brown)("Determine "x_("Intercepts"))

Set $y$ to 0 giving

$0 = {\left(x + 5\right)}^{2} - 3$

${\left(x + 5\right)}^{2} = + 3$

Square root both sides

$x + 5 = \pm \sqrt{3}$

Subtract 5 from both sides

Exact values$\text{ } x = - 5 \pm \sqrt{3}$

Approximate values$\text{ "x=-6.73" and } x = - 3.27$ to 2 decimal places