How do write standard form of equation of the parabola given that vertex is (5/2, - 3/4) and graph passes through point (-2 ,4)?

1 Answer
Ken C.
Aug 23, 2016

Express the parabola in vertex form and work from there to obtain y=1981x2−9581x+5881.

Explanation:

We might just want to jump into the problem and try finding the standard form y=ax2+bx+c of this parabola, but we don't have enough information to do this yet. Instead, we have to use the vertex form of a parabola:
y=a(x−h)2+k
Where (h,k) is the vertex.

The vertex is given in the problem, so:
y=a(x−52)2−34

We have to find the value of the constant a (which determines how thin or thick the parabola is), and we can do this using the point (−2,4):
y=a(x−52)2−34
→4=a(−2−52)2−34
→4+34=a(−92)2
→194=814a
→a=1981

Our equation in vertex form is therefore:
y=1981(x−52)2−34

We can expand the right hand side to find the standard form:
y=1981(x−52)2−34
→y=1981(x2−5x+254)−34
→y=1981(x2−5x+254)−34
→y=1981x2−9581x+475324−34
→y=1981x2−9581x+5881

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