Answer:
3) [tex]y=(x+3)^2+2[/tex]
Step-by-step explanation:
The vertex for a quadratic equation is given by:
[tex]y=a(x-h)^2+k[/tex]
Where [tex]a[/tex] ≠0 and [tex](h,k)[/tex] is the vertex of the the curve.
From the graph we see that the vertex is [tex](-3,2)[/tex]
Thus the equation of the graph can be written as:
[tex]y=a(x-(-3))^2+2[/tex]
[tex]y=a(x+3)^2+2[/tex]
Given y-intercept on the graph [tex](0,11)[/tex]
Plugging in point of y-intercept in the equation we got in order to find [tex]a[/tex]
[tex]11=a(0+3)^2+2[/tex]
[tex]11=a(3)^2+2[/tex]
[tex]11=9a+2[/tex]
Subtracting both sides by 2.
[tex]11-2=9a+2-2[/tex]
[tex]9=9a[/tex]
Dividing both sides by 9.
[tex]\frac{9}{9}=\frac{9a}{9}[/tex]
∴ [tex]a=1[/tex]
So the equation of graph is
[tex]y=(1)(x+3)^2+2[/tex]
[tex]y=(x+3)^2+2[/tex]
