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

Answer:

1)  $y=x^2+2$

2)  $y=x^2-4$

3)  $y=x^2$

4)  $y=x^2+6x+14$

Step-by-step 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 $x+3$:

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

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