Question

PLEASE HELPPP!!!!

1. ) The points (-2, 6), (-1, 3), (0, 2), and (1, 3) are on the graph of my function. What is the equation?

2. ) My parabola intersects the x-axis at points (-2, 0) and (2, 0) and the y-axis at point (0, -4). What is the equation?

3.) The table of values below describe the relationship for my function.

x values: -2, -1, 0 , 1
y values: 4, 1, 0, 1

What is the equation?

4.) To find my equation, shift f(x) = x2 + 5 left 3.

What is the equation?

Answer 1

1)
`y=x^2+2`

2)
`y=x^2-4`

3)
`y=x^2`

4)
`y=x^2+6x+14`

Explanation:

Question 1

Given points: (-2, 6) (-1, 3) (0, 2) (1, 3)

This is likely to be a quadratic equation with a vertex of (0, 2) since the points (-1, 3) and (1, 3) are symmetrical about x = 0

Vertex form of quadratic equation:
`y=a(x-h)^2+k`

(where (h, k) is the vertex)

`\implies y=a(x-0)^2+2`

`\implies y=ax^2+2`

To find a, input one of the points into the equation:

`(1, 3)\implies a(1)^2+2=3\implies a=1`

Therefore, the equation of the graph is
`y=x^2+2`

Question 2

If the parabola intersects the x-axis at (-2, 0) and (2, 0) then

`y=a(x+2)(x-2)`

We are told that the y-intercepts if (0, -4). Input this into the equation to find a:

`\implies a(0+2)(0-2)=-4`

`\implies -4a=-4`

`\implies a=1`

Therefore,

`\implies y=(x+2)(x-2)`

`\implies y=x^2-4`

Question 3

Given points: (-2, 4) (-1, 1) (0, 0) (1, 1)

This is likely to be a quadratic equation with a vertex of (0, 0) since the points (-1, 1) and (1, 1) are symmetrical about x = 0

Vertex form of quadratic equation:
`y=a(x-h)^2+k`

(where (h, k) is the vertex)

`\implies y=a(x-0)^2+0`

`\implies y=ax^2`

To find a, input one of the points into the equation:

`(-2,4)\implies a(-2)^2=4\implies a=1`

Therefore, the equation of the graph is
`y=x^2`

Question 4

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

If we wish to move the function to the left by 3 units, we substitute x for
`x+3`:

`\begin{aligned}\implies f(x+3) & =(x+3)^2+5\\ & = x^2+6x+9+5\\ & = x^2+6x+14\end{aligned}`

`\implies y=(x+3)^2+5`