Question

What is the equation of the quadratic function that has vertex (-3, 1) and passes through the point (-2, 4)?

Answer 1

• Vertex form: $y=a(x-h)^2+k$ , with (h,k) as the vertex.

For this, we will be using vertex form. Firstly, plug the vertex into the vertex form equation:

$y=a(x-(-3))^2+1\\y=a(x+3)^2+1$

Next, we need to solve for a. Plug in (-2,4) into the x and y coordinates to solve for a as such:

$4=a(-2+3)^2+1\\4=a(1)^2+1\\4=a+1\\3=a$

Putting our equation together, it's $y=3(x+3)^2+1$

*Additional section*

• Standard form: $y=ax^2+bx+c$

Converting to standard form as such:

$y=3(x+3)^2+1\\y=3(x^2+6x+9)+1\\y=3x^2+18x+27+1\\y=3x^2+18x+28$