Question

Find two points on the graph of the parabola other than the vertex and x-intercepts

Answer 1

(0, -5) and (2, -8.75)

Explanations

Given the function expressed as:

`g(x)=(x-2)^2-9`

We are to find two other coordinates other than the vertex and x-intercept. Note that the equation of a parabola in vertex form is expressed as:

`y=a\mleft(x-h\mright)^2+k`

Another coordinate point is the y-intercept of the graph. The y-intercept occurs at the point where x = 0.

Substitute x = 0 into the given function;

`\begin{gathered} g(x)=(x-2)^2-9 \\ g(0)=(0-2)^2-9 \\ g(0)=(-2)^2-9 \\ g(0)=4-9 \\ g(0)=-5 \end{gathered}`

Hence the y-intercept of the graph is (0, -5)

The other coordinate of the function we can determine is the FOCUS.

The coordinates for the focus of the parabola is given as (h, k+1/4a).

where:

• a is the intercept

,

• (h, k) is the vertex

Comparing the given function with the general function, we can see that:

a = 1

h = 2

k = -9

Substitute these values into the coordinate of the focus will give:

`\begin{gathered} \text{Focus}=(h,k+(1)/(4a)) \\ \text{Focus}=(2,-9+(1)/(4(1))) \\ \text{Focus}=(2,(-36+1)/(4)) \\ \text{Focus}=(2,-8.75) \end{gathered}`

Therefore the other two points on the graph are (0, -5) and (2, -8.75)