Equation of Parabola Problem

Question

A) Determine the equation of the parabola with a vertex at (-2, 12) and also goes through (1, -15). Express your answer in both vertex form, as well as general (ax2+bx+c) form.

B) Solve the quadratic equation 3x2−6𝑥+1=16

C) Solve (3𝑥+7)2=121.

Answer 1

You should try some of these yourself. It is not the job of a tutor to do your homework for you.

If you cannot do some of the problems, ask for specific help; if you cannot do any, you need a tutor.

a) The vertex form of the parabola is A(x-a)² + b where a is the abscissa of the vertex and b is the ordinate.

A can be determined from the 2nd point. In this case the answers are a=-2, b=12 and A = -3.

b) Use the quadratic formula.

c) As written this is a simple linear equation. If the "2" is supposed to be an exponent, then again use the quadratic formula.

Answer 2

1). Vertex form would give you y=a(x+2)^2 + 12

Use the given point(1,-15) to solve for a

-15 = a(1+2)^2 +12

-27 = 3a. A=-3 so equation in vertex form is y=-3(x+2)^2 +12

Expand this to get standard form y= -3(x^2 +4x+ 4)+12

y= -3x^2 -12x -12 +12

y= -3x^2 -12x

2) 3x^2 - 6x + 1 = 16 Set to zero 3x^2 - 6x -15 =0 divide both sides by 3

x^2 -2x -5 = 0 Since this does not factor can use quadratic equation or completing the square to solve for x I used completing the square. (X-1)^2 -5 -1=0

(x-1)^2= 6 X-1= +/-square root 6

x= 1+/- square root 6

3) (3x+7)^2 =121. Take the square root of both sides 3x+ 7= +/-11

3x= -18 x=-6

3x= 4 x=4/3