Solve for b f(x)=1/4x^2+bx+10

$f (x) = \frac{1}{4} x^{2} + b x + 10$

Step 1

Rewrite the equation as $\frac{1}{4} x^{2} + b x + 10 = f (x)$ .

$\frac{1}{4} x^{2} + b x + 10 = f (x)$

Step 2

Combine $\frac{1}{4}$ and $x^{2}$ .

$\frac{x^{2}}{4} + b x + 10 = f (x)$

Step 3

Move all terms not containing $b$ to the right side of the equation.

Subtract $\frac{x^{2}}{4}$ from both sides of the equation.

$b x + 10 = f (x) - \frac{x^{2}}{4}$

Subtract $10$ from both sides of the equation.

$b x = f (x) - \frac{x^{2}}{4} - 10$

Step 4

Divide each term in $b x = f (x) - \frac{x^{2}}{4} - 10$ by $x$ and simplify.

Divide each term in $b x = f (x) - \frac{x^{2}}{4} - 10$ by $x$ .

$\frac{b x}{x} = \frac{f (x)}{x} + \frac{- \frac{x^{2}}{4}}{x} + \frac{- 10}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{b x}{x} = \frac{f (x)}{x} + \frac{- \frac{x^{2}}{4}}{x} + \frac{- 10}{x}$

Divide $b$ by $1$ .

$b = \frac{f (x)}{x} + \frac{- \frac{x^{2}}{4}}{x} + \frac{- 10}{x}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$b = \frac{f (x)}{x} - \frac{x^{2}}{4} \cdot \frac{1}{x} + \frac{- 10}{x}$

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{x^{2}}{4}$ into the numerator.

$b = \frac{f (x)}{x} + \frac{- x^{2}}{4} \cdot \frac{1}{x} + \frac{- 10}{x}$

Factor $x$ out of $- x^{2}$ .

$b = \frac{f (x)}{x} + \frac{x (- x)}{4} \cdot \frac{1}{x} + \frac{- 10}{x}$

Cancel the common factor.

$b = \frac{f (x)}{x} + \frac{x (- x)}{4} \cdot \frac{1}{x} + \frac{- 10}{x}$

Rewrite the expression.

$b = \frac{f (x)}{x} + \frac{- x}{4} + \frac{- 10}{x}$

Move the negative in front of the fraction.

$b = \frac{f (x)}{x} - \frac{x}{4} + \frac{- 10}{x}$

Move the negative in front of the fraction.

$b = \frac{f (x)}{x} - \frac{x}{4} - \frac{10}{x}$