Question

Write an equation of the parabola that passes through the point (-9, 2) and has vertex (-3,1).

Answer 1

`\textsf{Vertex form}: \quad y=(1)/(36)(x+3)^2+1`

`\textsf{Standard form}: \quad y=(1)/(36)x^2+(1)/(6)x+(5)/(4)`

Explanation:

`\boxed{\begin{minipage}{5.6 cm}\underline{Vertex form of a quadratic equation}\\\\$y=a(x-h)^2+k$\\\\where:\\ \phantom{ww}$\bullet$ $(h,k)$ is the vertex. \\ \phantom{ww}$\bullet$ $a$ is some constant.\\\end{minipage}}`

Given:

• Vertex = (-3, 1)
• Point = (-9, 2)

Substitute the given vertex and point into the formula and solve for a:

`\begin{aligned}y&=a(x-h)^2+k\\\\\implies 2&=a(-9-(-3))^2+1\\2&=a(-6)^2+1\\2&=36a+1\\36a&=1\\a&=(1)/(36)\end{aligned}`

Therefore, the equation of the parabola in vertex form is:

`\boxed{y=(1)/(36)(x+3)^2+1}`

In standard form:

`\implies y=(1)/(36)(x+3)(x+3)+1`

`\implies y=(1)/(36)(x^2+6x+9)+1`

`\implies y=(1)/(36)x^2+(6)/(36)x+(9)/(36)+1`

`\implies y=(1)/(36)x^2+(1)/(6)x+(5)/(4)`