Question

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

Answer 1

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+10x+25-3`, i.e., `y+3=(x+5)^2`. Hence vertex (-5,-3), etc.

Answer 2

Detailed solution using completing the square

`color(blue)("Vertex "->(x,y)->(-5,-3))`

`color(blue)("Exact values "x_("intercepts")=-5+-sqrt(3)`

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

Explanation:

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

`color(brown)("Move the power from "x^2" to outside the brackets")`

`y=(x+10x)^2+22 + k`

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

`y=(x+10)^2+22 + k`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Multiply the 10 by "1/2)`

`y=(x+5)^2+22 + k` ...............................Equation (2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
`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" "->" "k=-25`

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

So equation (2) becomes

`y=(x+5)^2-3`.............................Equation (3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(brown)("Reading directly from Equation (3) "larr "vertex")`

`x_("vertex")=(-1)xx5 = -5`

`y_("vertex")=-3`

`color(blue)("Vertex "->(x,y)->(-5,-3))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("Determine "x_("Intercepts"))`

Set `y` to 0 giving

`0=(x+5)^2-3`

Add 3 to both sides

`(x+5)^2=+3`

Square root both sides

`x+5 = +-sqrt(3)`

Subtract 5 from both sides

Exact values`" "x=-5+-sqrt(3)`

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