#### To determine

**To find:** The equation of parabola.

#### Answer

The equation of parabola is y=3x2−2x+7.

#### Explanation

**Given:**

The equation of the parabola y=ax2+bx+c (1)

The slope of the parabola are 4 at x=1 and -8 at x=−1.

The parabola passing through the point (2,15).

**Derivative rules:**

(1) Constant multiple rule: ddx(cf)=cddx(f)

(2) Power rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

**Calculation:**

Obtain the values of *a*, *b* and *c* in the equation of parabola.

Take the point (2,15).

Here, x=2 and y=15 .

Substitute 2 for *x* and 15 for y in equation (1).

15=a(2)2+b(2)+c(2)15=a(4)+b(2)+c(2)15=4a+2b+c

Therefore, the equation is 4a+2b+c=15 (2)

Obtain the derivative of y=ax2+bx+c.

y′(x)=ddx(ax2+bx+c)

Apply the sum rule (3) and the constant multiple rule (1),

y′(x)=ddx(ax2)+ddx(bx)+ddx(c)=addx(x2)+bddx(x)+ddx(c)

Since the derivative of constant function is zero, then ddx(c)=0.

y′(x)=addx(x2)+bddx(x)+0=addx(x2)+bddx(x)

Apply power rule (2) and simplify the expression,

y′(x)=a(2x2−1)+b(x1−1)=a(2x1)+b(x0)=2ax+b

Therefore, the first derivative of y is y′(x)=2ax+b.

Note that, the parabola has slope 4 at x=1,−8,−1.

That is,y′(1)=4 and y′(−1)=−8.

Substitute 1 for *x* in y′(x)=2ax+b, and use y′(1)=4 to get,

y′(1)=2a+b4=2a+b

Therefore, the equation is 2a+b=4 (3)

Substitute −1 for *x* in y′(x)=2ax+b and use y′(−1)=−8 to get,

y′(−1)=−2a+b−8=−2a+b

Therefore, the equation is .−2a+b=−8 (4)

Add equation (3) and (4),

2b=−4b=−2

Thus, the value of b=−2.

Substitute -2 for *b* in equation (3),

2a−2=42a=4+22a=6a=3

Thus, the value of a=3.

Substitute 3 for *a* and -2 for *b* in equation (2),

4(3)+2(−2)+c=1512−4+c=158+c=15c=7

Thus, the value of c=7.

Substitute the value of *a*,*b*, and *c* in equation (1),

y=3x2+(−2)x+7

Therefore the equation of parabola is y=3x2−2x+7.