Question

How do you find the x and y intercepts for `y = -|x+10|`?

Answer 1

General shape is `^^^`

`y_("intercept")->(x,y)=(0,-10)`

`x_("intercept")->(x,y)=(10,0)`

Vertex `->(x,y)=(-10,0)`

Explanation:

`color(blue)("Tip 1: Shape of the graph")`

If an absolute is positive we get the shape `vvv`

If an absolute is negative we get the shape `^^^color(red)( larr" Our one")`

This follows the same pattern as with a quadratic.

If the `x^2` term is positive we get `uuu`

If the `x^2` term is negative we get `nnn`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Tip 2: horizontal position")`

If you add a value to the `x` then it moves the graph left
If you subtract a value from the `x` then it moves the graph right

Example: suppose we had say

`y=x^2+2x-2`
then we dicide to change it so that we add 4 to `x`. We have:

`y=(x+4)^2+2(x+4)-2`

Because we have added 4 it moves the graph `y=x^2_2x-2` left by 4

`color(red)("Our one:")`
So if we add 10 to `y=-|x|` giving `y=-|x+10|` we move the graph of `y=-|x|` left by 10
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Tip 3: x-intercept")`

The graph crosses the x-axis at `y=0`

`y=-|x+10| color(white)("d")-> color(white)("d") 0=-|x-10|`

The only way we can obtain 0 as the value of `y` is if `x=+10`

So we have: `x_("intercept")->(x,y)=(10,0)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Tip 4: y-intercept")`

The graph crosses the y-axis at `x=0`

`y=-|0+10| = -10`

So we have: `y_("intercept")->(x,y)=(0,-10)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Tip 4: The vertex")`

This occurs when the overall value `color(red)("within")` the absolute is about to 'flip' sign from positive to negative. That is, it becomes 0 which happens at `x=-10`

`y=-|color(magenta)(x)+10|`

`color(limegreen)(y)=-|ubrace(color(magenta)(-10)+10)| = color(limegreen)(0)`
`color(white)("ddddddddd")darr`
`color(white)("dddddddd.d")0`

Vertex `->(x,y)=(color(magenta)(-10),color(limegreen)(0))`