Question

Number 11. Find a quadratic equation with (-2,3) and y intercept of 11

Answer 1

`y=2x^2+8x+11`

Step-by-step explanation:

A quadratic equation in vertex form is generally given as;

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

where (h, k) is the coordinate of the vertex.

Given the coordinate (-2, 3), we'll have that;

h = -2

k = 3

Given a y-intercept of 11 and we know that at the y-intercept x = 0.

Substituting the above values into the vertex form equation and solving for a, we'll have;

`\begin{gathered} 11=a\lbrack0-(-2)\rbrack^2+3 \\ 11=4a+3 \\ 4a=8 \\ a=(8)/(4) \\ a=2 \end{gathered}`

Substituting a = 2, h = -2 and k = 3 into the vertex form equation and taking it to standard form, we'll have;

`\begin{gathered} y=2(x+2)^2+3 \\ y=2(x^2+4x+4)+3 \\ y=2x^2+8x+8+3 \\ y=2x^2+8x+11 \end{gathered}`