Find f (-2) f(x)= 3x^2 - 8x + 7

Solving for f. Want to solve for x instead?

1 Remove parentheses.

f\times -2fx=3{x}^{2}-8x+7f×−2fx=3x

2

−8x+7

2 Use Product Rule: {x}^{a}{x}^{b}={x}^{a+b}x

a

x

b

=x

a+b

.

-{f}^{2}\times 2x=3{x}^{2}-8x+7−f

2

×2x=3x

2

−8x+7

3 Regroup terms.

-2{f}^{2}x=3{x}^{2}-8x+7−2f

2

x=3x

2

−8x+7

4 Divide both sides by -2−2.

{f}^{2}x=-\frac{3{x}^{2}-8x+7}{2}f

2

x=−

2

3x

2

−8x+7

5 Divide both sides by xx.

{f}^{2}=-\frac{\frac{3{x}^{2}-8x+7}{2}}{x}f

2

=−

x

2

3x

2

−8x+7

6 Simplify  \frac{\frac{3{x}^{2}-8x+7}{2}}{x}

x

2

3x

2

−8x+7

  to  \frac{3{x}^{2}-8x+7}{2x}

2x

3x

2

−8x+7

.

{f}^{2}=-\frac{3{x}^{2}-8x+7}{2x}f

2

=−

2x

3x

2

−8x+7

7 Take the square root of both sides.

f=\pm \sqrt{-\frac{3{x}^{2}-8x+7}{2x}}f=±√

−

2x

3x

2

−8x+7

8 Simplify  \sqrt{-\frac{3{x}^{2}-8x+7}{2x}}√

−

2x

3x

2

−8x+7

  to  \sqrt{\frac{3{x}^{2}-8x+7}{2x}}\imath√

2x

3x

2

−8x+7

ı.

f=\pm \sqrt{\frac{3{x}^{2}-8x+7}{2x}}\imathf=±√

2x

3x

2

−8x+7

ı

9 Regroup terms.

f=\pm \imath \sqrt{\frac{3{x}^{2}-8x+7}{2x}}f=±ı√

2x

3x

2

−8x+7

Done- :)

f=±ı√ 2x 3x 2 −8x+7

Step-by-step explanation

ian gone lie this got me confused

step-by-step explanation:

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents