Question

NO LINKS!!! Find the equation of the PARABOLA with a vertex at (-2, 6) and passing through the point (1, -3)​

Answer 1

General equation of parabola

• y=a(x-h)²+k

for vertex(h,k)

Now

`\\ \rm\Rrightarrow y=a(x+2)^2+6`

• Put (1,-3) and find a

`\\ \rm\Rrightarrow -3=a(1+2)^2+6`

`\\ \rm\Rrightarrow -9=a(3)^2`

`\\ \rm\Rrightarrow -9=9a`

`\\ \rm\Rrightarrow a=-1`

So

equation of parabola

`\\ \rm\Rrightarrow y=-(x+2)^2+6`

Answer 2

`y=-(x+2)^2+6`

Explanation:

Vertex form of a quadratic equation

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

where:

• (h, k) is the vertex
• a is some constant

Given:

• vertex = (-2, 6)
• point on parabola = (1, -3)

Substitute the given values into the vertex equation and solve for a:

`\implies -3=a(1-(-2)^2+6`

`\implies -3=a(3)^2+6`

`\implies -3=9a+6`

`\implies 9a=-9`

`\implies a=-1`

Vertex form

Substitute the given vertex and the found value of a into the vertex equation:

`\implies y=-(x+2)^2+6`

Standard form

Expand the brackets of the vertex form:

`\implies y=-(x^2+4x+4)+6`

`\implies y=-x^2-4x+2`