Question

8.

Answer 1

Answer:    y= 4x² +2x -4

C. y = 4x2 + 2x − 4

Step-by-step explanation:  f(x)=4x^2+2x-4

Given three points

P1(-2,8)

P2(0,-4)

P3(4,68)

We need the quadratic equation that passes through all three points.

Solution:

We first assume the final equation to be

f(x)=ax^2+bx+c .............................(0)

Observations:

1. Points are not symmetric, so cannot find vertex visually.

2. Using the point (0,-4) we substitute x=0 into f(x) to get

f(0)=0+0+c=-4, hence c=-4.

3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.

f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)

f(4)=a(4^2)+b(4)-4=68 =>  16a+4b-4=68.............(2)

4. Solve system

2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)

substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2  ..........(4)

5. Put values c=-4, a=4, b=2 into equation (0) to get

f(x)=4x^2+2x-4

Check:

f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8

f(0)=0+0-4 = -4

f(4)=4(4^2)+2(4)-4=64+8-4=68

So all consistent, => solution ok.

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Answer:

y = 4x² + 2x − 4

Step-by-step explanation:

The y-values go down and up again, so the parabola opens upward. That means the x² coefficient must be positive. All answer choices are eliminated except one.

y = 4x² +2x -4

 There are other ways to get the answer. One of them is to use the quadratic regression function of a spreadsheet or graphing calculator. The one attached shows the points are matched by the equation ...

y = 4x² +2x -4 . . . . . as above

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