Answer:
[tex]y=-(x+2)^2+6[/tex]
Step-by-step explanation:
Vertex form of a quadratic equation
[tex]y=a(x-h)^2+k[/tex]
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:
[tex]\implies -3=a(1-(-2)^2+6[/tex]
[tex]\implies -3=a(3)^2+6[/tex]
[tex]\implies -3=9a+6[/tex]
[tex]\implies 9a=-9[/tex]
[tex]\implies a=-1[/tex]
Vertex form
Substitute the given vertex and the found value of a into the vertex equation:
[tex]\implies y=-(x+2)^2+6[/tex]
Standard form
Expand the brackets of the vertex form:
[tex]\implies y=-(x^2+4x+4)+6[/tex]
[tex]\implies y=-x^2-4x+2[/tex]
