(algebra 1) Sam graphs a quadratic function m(x) which has a vertex at (-5,10) and an y-intercept at -15. Write a function in vertex form to represent m(x)
Would appreciate the help.

Question

(algebra 1) Sam graphs a quadratic function m(x) which has a vertex at (-5,10) and an y-intercept at -15. Write a function in vertex form to represent m(x)    
Would appreciate the help.

ThanosQC · Answer

Vertex form:
f(x) = a(x - h)^2 + k, where h & k are the points on the vertex
so
f(x) = a(x+5)^2 + 10
To solve for a, replace f(x) and x with the y-intercept's coordinates, (0, -15) and solve for a
-15 = a(0+5)^2 + 10
-15 = a(5)^2 + 10
-15-10 = 25a + 10-10
25a/25 = -25/25
a = -1
Now you know a and can write the completed function:
f(x) = -1(x+5)^2 + 10

gelphielvr · Answer

@thanosqc wrote:

Vertex form: f(x) = a(x - h)^2 + k, where h & k are the points on the vertex so f(x) = a(x+5)^2 + 10 To solve for a, replace f(x) and x with the y-intercept's coordinates, (0, -15) and solve for a -15 = a(0+5)^2 + 10 -15 = a(5)^2 + 10 -15-10 = 25a + 10-10 25a/25 = -25/25 a = -1 Now you know a and can write the completed function: f(x) = -1(x+5)^2 + 10

thank u Thanos :)) cool pfp btw

ThanosQC · Answer

@gelphielvr wrote:

@thanosqc wrote:

Vertex form: f(x) = a(x - h)^2 + k, where h & k are the points on the vertex so f(x) = a(x+5)^2 + 10 To solve for a, replace f(x) and x with the y-intercept's coordinates, (0, -15) and solve for a -15 = a(0+5)^2 + 10 -15 = a(5)^2 + 10 -15-10 = 25a + 10-10 25a/25 = -25/25 a = -1 Now you know a and can write the completed function: f(x) = -1(x+5)^2 + 10

thank u Thanos :)) cool pfp btw

Yw and thank you!