SOLUTION: Find the equation of the quadratic curve with a turning point at (-2,3) and which passes through (-1,5)

Algebra ->  Test -> SOLUTION: Find the equation of the quadratic curve with a turning point at (-2,3) and which passes through (-1,5)      Log On


   



Question 1184796: Find the equation of the quadratic curve with a turning point at (-2,3) and which passes through (-1,5)
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Find the equation of the quadratic curve with
if a turning point at (-2,3), that is a vertex: h=-2 and k=3
so use equation in vertex form
y=a%28x-h%29%5E2%2Bk
y=a%28x-%28-2%29%29%5E2%2B3
y=a%28x%2B2%29%5E2%2B3

and since passes through (-1,5), use it to calculate a
5=a%28-1%2B2%29%5E2%2B3
5=a%281%29%5E2%2B3
5=a%2B3
a=5-3
a=2
your equation is:

y=2%28x%2B2%29%5E2%2B3











Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


One form of the general vertex form of the the equation of a parabola is

y-k=a%28x-h%29%5E2

The vertex is given as (-2,3), so an equation of the parabola is

y-3=a%28x%2B2%29%5E2

The solution from the other tutor uses the other point (-1,5) and formal algebra to determine the coefficient a; that is of course fine.

Once you become familiar with quadratic equations you can find the coefficient a informally, as follows:

The vertex is (-2,3) and the other given point is (-1,5). So 1 unit away from the vertex in the x direction the y value has increased by 2. That means the coefficient a is 2. So

ANSWER: y-3=2%28x%2B2%29%5E2