Question

How do you find the vertex and intercepts for `(x+3)^2=16(y-2)`?

Answer 1

Vertex `->(x,y)=(-3,2)`

y-intercpt `->(0,41/16)`

x-intercept `->` None

Explanation:

This is almost in 'completed square' form.

Sometimes called vertex form.

We need to get the `y` on its own.

`y-2=1/16(x+3)^2`

`y=1/16(xcolor(red)(+3))^2color(blue)(+2) larr" Now it is in vertex form"`
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`color(green)("To determine the vertex")`

`x_("vertex")=(-1)xx(color(red)(+3)) = -3`

`y_("vertex") =color(blue)(+2)`

Vertex `->(x,y)=(-3,2)`
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`color(green)("To determine the " y" intercept")`

This occurs at `x=0` so by substitution:

`y=1/16(x+3)^2+2" "->" "y=9/16+2 =41/16`

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`color(green)("To determine the " x" intercept")`

This occurs at `y=0` so by substitution:

`y=1/16(x+3)^2+2" "->" "-2=1/16(x+3)^2`

`color(white)("mmmmmmmmmmmb")->" "-32=(x+3)^2`

Square root both sides

`sqrt(-32)=x+3 color(red)(larr" This result has a problem")`

You can take the square root of a negative number but the answer is in the realm of Complex Numbers. Consequently the graph does not cross the x-axis

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Foot Note: The `x^2` term is positive so the graph is of shape `uu`.

The vertex (which is a minimum value in this case) is above the x-axis. That alone tells you that there is no solution to `y=0` that is in the set of 'Real Numbers'. That is `y!=0" for " x inRR`