Answer:
Step-by-step explanation:
If you plot the focus and the vertex, you see that the focus is on the same vertical line, just 4 units up. Since the vertex is below the focus and a parabola ALWAYS wraps around the focus, the parabola is a positive y equals x-squared type. Depending upon what you call standard form will dictate how your answer "looks", although they are both the same parabola. There are 2 forms:
[tex](x-h)^2=4p(y-k)[/tex] and
[tex]ax^2+bx+c=y[/tex]
We will work on the first one, then rewrite it into the second one.
Our h value is -5, our k value is 2 (from the vertex (h,k)), and p is defined as the distance between the focus and the vertex. Our p is 4. Filling all that in:
[tex](x+5)^2=16(y-2)[/tex]
That's one form. If we expand the left side of that form we have
[tex]x^2+10x+25=16(y-2)[/tex] Now divide both sides by 16 to get
[tex]\frac{1}{16}x^2+\frac{5}{8}x+\frac{25}{16}=y-2[/tex]
Now add 2 to both sides in the form 32/16 to get
[tex]\frac{1}{16}x^2+\frac{5}{8}x+\frac{57}{16}=y[/tex]
They are both the same parabola; pick whichever one fits your needs.
