Question

What is the equation, in standard form, of a parabola that contains the following points?. . (–2, 18), (0, 2), (4, 42). . A) y = –2x^2 – 2x – 3. B) y = –3x^2 + 2x – 2. C) y = 3x^2 – 2x + 2. D) y = –2x^2 + 3x + 3

Answer 1

Hello,
Answer C
y=ax²+bx+c
(0,2)==>2=a*0²+b*0+c==>c=2

(4,42)==>42=a*16+4b+2==>4a+b=10 (1)
(-2,18)==>18=a*4-2*b+2==>4a-2b=16 (2)

(1)-(2) ==>3b=10-16 ==>b=-2
4a+(-2)=10==>4a=12==>a=3

y=3x²-2x+2 is the equation

Answer 2

The equation of parabola is:
y = ax² + bx + c.

consider the (0, 2) to get the intercept.
substitute x = 0 and y = 2 into the equation.
2 = a(0)² + b(0) + c
c = 2
Similarly,
considering the point (- 2, 18):
Substitute x = - 2 ,y = 18 ,and c = 2
18 = a(- 2)² + b(- 2) + 2
18 = 4a - 2b + 2
Simplifying:
4a - 2b = 16
2a - b = 8
Solve for b:
Hence,
b = 2a - 8 ------(1)

Now,
Substitute x = 4,y = 42, and c = 2
42 = a(4)² + b(4) + 2
42 = 16a + 4b + 2
16a + 4b = 40
divide each term by 4.
We get,
4a + b = 10 -------(2)

Substitute b= 2a - 8 in Equation (2),
We get:
4a + 2a - 8 = 10
6a = 18 + 8
6a = 18
a = 3
Substituting a = 3 in b= 2a - 8
b = 6-8
b = -2
Hence,
our final equation will be:
y = 3x² - 2x + 2