Studying for my Grade 10 Math exam I found a question on quadratic formulas that I can't remember. It shows a parabola with the Vertex $( -3, -3)$ and a point marked $(-1,1)$.
The question asks me to algebraically solve the $a$ value in $y=a(x-h)^2+k.$
I have gotten as far as $1=a(-1+3)-2$ but none of my answers have made sense according to the graph and am unsure of where to go from here. Maybe I'm plugging it in wrong or something but I would appreciate some input.
On
I believe you mean $y=a(x-h)^2+k.$ Since we know that $(-3,3)$ is the vertex, $x=-3$ is the line of symmetry so $y=a(x+3)^2+k.$ We plug $x=-3$ in and since $a(x+3)^2+k=0,$ $k=-3.$ Now that we know $k=-3,$ we plug the equation in for the point. $1=a(-1+3)-3$ so $1=2a-3.$ Solving this we get $a=2.$ The equation is $\boxed{y=1(x-(-3))^2-3}$ which simplified is $\boxed{y=(x+3)^2-3}.$
Assuming $(-1, 1)$ is a point on the parabola, and with your equation of the form $$y=a(x-h)^2+k$$ where $(-3,-3) = (h, k)$, that will give you $$y=a(x-(-3))^2 + (-3) \iff y=a(x+3)^2 -3\tag{$\dagger$}$$
Now substitute $x=-1$ and $y=1$ into the variables x, y, (since you know $(-1, 1)$ lies on the parabola, of that equation ($\dagger$)), and solve for $a$:
$$\underbrace{y}_{1}=a(\underbrace{x}_{-1}+3)^2-3$$
Once you find the value of $a$, substitute that value where $a$ occurs in the equation ($\dagger$).