Question

The table describes the quadratic function h(x).


x h(x)
−3 1
−2 −2
−1 −3
0 −2
1 1
2 6
3 13

What is the equation of h(x) in vertex form?

Answer 1

To find the equation of h(x) in vertex form, we need to complete the square. First, we can find the axis of symmetry by averaging the x-values of the two points on either side of the vertex:

Axis of symmetry = (−2 + (−1))/2 = −1.5

This tells us that the x-coordinate of the vertex is −1.5. To find the y-coordinate, we can substitute this value into the function:

h(−1.5) = −3

So the vertex is at (−1.5, −3). Now we can write the equation in vertex form:

h(x) = a(x − (−1.5))2 − 3

To find the value of a, we can use one of the other points on the graph. Let's use (2, 6):

6 = a(2 − (−1.5))2 − 3
9 = 5.25a
a = 9/5.25 = 1.71 (rounded to two decimal places)

So the equation of h(x) in vertex form is:

h(x) = 1.71(x + 1.5)2 − 3

I hope this helps!

Answer 2

let \[h(x)=ax^2+bx+c~~ ...(1)\]
when x=0, h(x)=-2
put in (1)
\[-2=a(0)^2+b(0)+c\]
c=-2
\[h(x)=ax^2+bx-2~~...(2)\]
when x=-1,h(x)=-3
\[-3=a(-1)^2+b(-1)-2\]
-3=a-b-2
a-b=-3+2=-1
a-b=-1 ...(3)
when x=1,h(x)=1
\[1=a(1)^2+b(1)-2\]
a+b=1+2
a+b=3 ...(4)
add (3) and (4)
2a=-1+3
2a=2
a=1
from (4)
1+b=3
b=3-1=2
so function is \[h(x)=x^2+2x-2\]
\[h(x)=x^2+2x+1-1-2\]
\[h(x)=(x+1)^2-3\]
which is in vertex form and vertex is (-1,-3)

Answer 3

jeez

Answer 4

idk