Write the equation of the parabola in vertex form. The vertex is (−3, 3) and the parabola passes through the point (4, 150). What is the equation for this?

Answer 1

Hi, Ella!

The vertex form of the equation is y=a(x-h)² + k

You know that the vertex (h,k) is (-3,3) and a point on the parabola is (4,150). So what you’re needed ing to find is a bc all of the other letters can be replaced into the equation.

First I would put the equation into vertex form using the vertex given in the problem. : y= (x - (-3))² + (3). Then I would use the ordered pair (4,150) and replace the x and y with those values

150 = a(4 + 3)² + 3

Now simplify

150 = a (7)² + 3

150 = 49a + 3 Important: please be sure to square. 🙂

147 = 49a

3= a

Next replace a into the formula now leaving x and y alone.

y = 3 (x+3)² + 3 [Vertex form]

I hope this helps. Have a great day!

Mrs. S

Answer 2

y = ax^2+bx+c

vertex is at -b/(2a) = -3

multiplies by -2a:

b = 6a

y = ax^2 + 6a x + c

(-3,3) --> 3 = 9a -18a + c

3 = c - 9a

3 + 9a = c

(4,150) --> 150 = 16a + 24a + c

150 = 40a + c

150 = 40a + 3 + 9a

150 = 49a + 3

147 = 49a

a=3

which forces c = 3+9(3) = 30

y = 3x^2 + 18x + 30

check: completes the square

y = 3x^2 + 18x + 30

y = 3(x^2+6x+10)

= 3(x^2+6x ) + 30

= 3(x^2+6x+9) + 30-27

= 3(x+3)^2 + 3

vertex is at (-3,3)

when x=4, y = 3(4+3)^2+3

= 3(7)^2 +3

= 3 (49)+3

= 3(50-1)+3

= 150-3+3

= 150

yes it checks

y = 3x^2 + 18x + 30

= 3(x+3)^2+3