Question

How do you find the vertex and the intercepts for `f(x) = 1/3(x+5)^2 + 5`?

Answer 1

Vertex`->(x,y)=(-5,5)`

`y_("intercept") = 13 1/3 ->40/3`

`x_("intercept")= =-5+[sqrt(15)] i" and "x=-5-[sqrt(15)] i`

Explanation:

Set `f(x)=1/3(x+5)^2+5 = y`................Equation(1)

`color(blue)("Determine the vertex")`
The given `y=1/3(x color(green)(+5))^2color(magenta)(+5)` is the vertex form equation.

You can almost directly read off the coordinate of the vertex from it.

`x_("vertex")=(-1)color(green)(xx5) = -5`
`y_("vertex")=color(magenta)(+5)`
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`color(blue)("Determine the y intercept")`

Set `x=0` giving:

`y_("intercept")=1/3(0+5)^2+5 = 13 1/3 ->40/3`
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`color(blue)("Determine the x intercept")`

From equation(1)

Set `1/3(x+5)^2+5=0`

`=>(x+5)^2=-15`

Square both sides

`x+5=+-sqrt(-15)`

As we have sqrt(-15) the graph does not cross the x-axis

So the only solution to `1/3(x+5)^2+5=0` is in the 'complex number' set of values.

`=>x=-5+-[sqrt(15)] i`

`=>x=-5+[sqrt(15)] i" and "x=-5-[sqrt(15)] i`

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