Given:
The vertex of a quadratic function is (-5,-1) and it passes through the point (-2,2).
To find:
The vertex and standard form of the quadratic function.
Solution:
The vertex form of a quadratic function is:
[tex]f(x)=a(x-h)^2+k[/tex]
Where, a is a constant, (h,k) is vertex.
The vertex of a quadratic function is (-5,-1). It means [tex]h=-5,k=-1[/tex].
[tex]f(x)=a(x-(-5))^2+(-1)[/tex]
[tex]f(x)=a(x+5)^2-1[/tex] ...(i)
The quadratic function passes through the point (-2,2). Putting [tex]x=-2,f(x)=2[/tex] in (i), we get
[tex]2=a(-2+5)^2-1[/tex]
[tex]2+1=a(3)^2[/tex]
[tex]3=9a[/tex]
[tex]\dfrac{1}{3}=a[/tex]
Putting [tex]a=\dfrac{1}{3}[/tex] in (i), we get
[tex]f(x)=\dfrac{1}{3}(x+5)^2-1[/tex]
Therefore, the vertex for of the quadratic function is [tex]f(x)=\dfrac{1}{3}(x+5)^2-1[/tex].
The standard form of a quadratic function is:
[tex]f(x)=Ax^2+Bx+C[/tex]
We have,
[tex]f(x)=\dfrac{1}{3}(x+5)^2-1[/tex]
[tex]f(x)=\dfrac{1}{3}(x^2+10x+25)-1[/tex]
[tex]f(x)=\dfrac{1}{3}x^2+\dfrac{10}{3}x+\dfrac{25}{3}-1[/tex]
[tex]f(x)=\dfrac{1}{3}x^2+\dfrac{10}{3}x+\dfrac{22}{3}[/tex]
Therefore, the standard form of a quadratic function is [tex]f(x)=\dfrac{1}{3}x^2+\dfrac{10}{3}x+\dfrac{22}{3}[/tex].