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

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

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Answer 1 · 2016-06-07T19:05:44

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

Explanation:

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

Answer 2 · 2016-06-07T19:52:39

Detailed solution using completing the square

$Vertex \to (x, y) \to (- 5, - 3)$ $color(blue)("Vertex "->(x,y)->(-5,-3))$

$Exact values x_{intercepts} = - 5 \pm \sqrt{3}$ $color(blue)("Exact values "x_("intercepts")=-5+-sqrt(3)$

Approx values $x = - 6.73 and x = - 3.27$ $" "x=-6.73" and " x=-3.27$ to 2 decimal places

Explanation:

Given: $y = x^{2} + 10 x + 22$ $" "y=x^2+10x+22$ ............................Equation (1)
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Write as $y = (x^{2} + 10 x) + 22 + k$ $" "y=(x^2+10x)+22 + k$

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

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

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

$Move the power from x^{2} to outside the brackets$ $color(brown)("Move the power from "x^2" to outside the brackets")$

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Remove the x from 10 x$ $color(brown)("Remove the "x" from "10x)$

$y = {(x + 10)}^{2} + 22 + k$ $y=(x+10)^2+22 + k$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Multiply the 10 by \frac{1}{2}$ $color(brown)("Multiply the 10 by "1/2)$

$y = {(x + 5)}^{2} + 22 + k$ $y=(x+5)^2+22 + k$ ...............................Equation (2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of k$ $color(brown)("Determine the value of "k)$

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

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$The completed square equation$ $color(brown)("The completed square equation")$

So equation (2) becomes

$y = {(x + 5)}^{2} - 3$ $y=(x+5)^2-3$ .............................Equation (3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$Reading directly from Equation (3) \leftarrow vertex$ $color(brown)("Reading directly from Equation (3) "larr "vertex")$

$x_{vertex} = (- 1) \times 5 = - 5$ $x_("vertex")=(-1)xx5 = -5$

$y_{vertex} = - 3$ $y_("vertex")=-3$

$Vertex \to (x, y) \to (- 5, - 3)$ $color(blue)("Vertex "->(x,y)->(-5,-3))$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine x_{Intercepts}$ $color(brown)("Determine "x_("Intercepts"))$

Set $y$ $y$ to 0 giving

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

Add 3 to both sides

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

Square root both sides

$x + 5 = \pm \sqrt{3}$ $x+5 = +-sqrt(3)$

Subtract 5 from both sides

Exact values $x = - 5 \pm \sqrt{3}$ $" "x=-5+-sqrt(3)$

Approximate values $x = - 6.73 and x = - 3.27$ $" "x=-6.73" and " x=-3.27$ to 2 decimal places

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

2 Answers

Explanation:

Explanation:

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