Precalculus-I · Unit 1 · Video 3 · Interactive Practice

Evaluating and Solving Functions Given by Formulas

IKey Formulas

Task	Formula	Answer
Evaluate	$f(2) = (2)^2 + 3(2) - 4 = 6$	Exactly one output
Solve	$h(p) = 3 \;\Rightarrow\; p^2 + 2p - 3 = 0$	Zero, one, or several inputs
Difference quotient	$\dfrac{f(a+h) - f(a)}{h},\ h \neq 0$	For $x^2 + 3x - 4$ : $2a + h + 3$
Function from an equation	$2n + 6p = 12 \;\Rightarrow\; p = 2 - \tfrac{1}{3}n$	Needs one output per input

Key Insight: $f(a + h)$ is not $f(a) + h$ : substitute $(a + h)$ for every $x$ , so every term containing the input changes.

IIEvaluate Forward, Solve Backward

For $h(p) = p^2 + 2p$ , an input gives exactly one output, but an output can come from two, one, or no inputs.

Evaluate at p 4 Solve h(p) = k 3

IIIThe Difference Quotient

For $f(x) = x^2 + 3x - 4$ , the simplified form $2a + h + 3$ equals $\dfrac{f(a+h) - f(a)}{h}$ for every $h \neq 0$ .

a 1 h 2

💡 The difference quotient is the slope of the line through $(a, f(a))$ and $(a+h, f(a+h))$ : rise $f(a+h) - f(a)$ over run $h$ .

IVWhen Does an Equation Define a Function?

An equation defines the output as a function of the input when each input gives exactly one output.

Equation Input 3

VQuiz Questions

Problem 1 · Evaluate, Then Solve

Given: $f(x) = x^2 - 4x$ .

What is $f(-1)$ ?

Which inputs give $f(x) = 5$ ?

✅ Correct! Evaluating gives the single output

f(-1) = 5

; solving

f(x) = 5

finds every input with that output,

-1

and

5

❌ Check the sign of $-4x$ . At

x = -1

it is

-4(-1) = +4

, not

-4

❌ Check the square.

(-1)^2 = 1

: the parentheses square the whole input, sign included.

❌ Two sign slips.

(-1)^2 = 1

and

-4(-1) = +4

❌ Not quite. Replace every

x

with

(-1)

, then calculate.

❌ Only half the answer.

f(5) = 5

, but solving means finding every input with output

5

, and there is another.

❌ Check the signs in the factors.

x^2 - 4x - 5 = (x - 5)(x + 1)

; test

x = 1

f(1) = -3 \neq 5

❌ Those solve $f(x) = 0$ . Set the formula equal to

5

first:

x^2 - 4x - 5 = 0

❌ Not quite. Set

x^2 - 4x = 5

, move the

5

across, and factor.

Show solution

Evaluate: replace every $x$ with $(-1)$ .

f(-1) = (-1)^2 - 4(-1) = 1 + 4 = 5

Solve: set the formula equal to $5$ .

x^2 - 4x = 5 \;\Rightarrow\; x^2 - 4x - 5 = 0 \;\Rightarrow\; (x - 5)(x + 1) = 0

So $x = 5$ or $x = -1$ . Check: $f(5) = 25 - 20 = 5$ ✓ and $f(-1) = 5$ ✓, the input from the first part.

Problem 2 · Substituting an Expression

Given: $f(x) = 3 - x^2$ — find $f(a + h)$ .

✅ Correct! The whole input

(a + h)

is squared, and the minus sign applies to every term of the square.

❌ The middle term is missing.

(a + h)^2 = a^2 + 2ah + h^2

, not

a^2 + h^2

❌ That is $f(a) + h$ , not $f(a + h)$ . Changing the input changes every term that contains it.

❌ The minus sign applies to the whole square.

-(a^2 + 2ah + h^2) = -a^2 - 2ah - h^2

❌ Not quite. Replace

x

with

(a + h)

in parentheses, then expand.

Show solution

Replace $x$ with $(a + h)$ , keeping the parentheses:

f(a + h) = 3 - (a + h)^2 = 3 - (a^2 + 2ah + h^2) = 3 - a^2 - 2ah - h^2

Compare $f(a) + h = 3 - a^2 + h$ : a different expression, because only the input should change.

Problem 3 · Simplify a Difference Quotient

Given: $f(x) = 2x^2 - x$ — simplify $\dfrac{f(a + h) - f(a)}{h}$ for $h \neq 0$ .

✅ Correct! Every term left in the numerator has a factor of

h

, which cancels the denominator.

❌ The $2$ multiplies the whole square.

2(a + h)^2 = 2a^2 + 4ah + 2h^2

, so

h^2

carries a

2

❌ That is only the numerator. Factor out

h

and cancel it against the denominator.

❌ Subtract all of $f(a)$ .

-(2a^2 - a) = -2a^2 + a

, so the

-a

terms cancel too.

❌ Not quite. Expand

f(a + h)

, subtract

f(a)

in parentheses, then factor out

h

Show solution

Step 1: substitute $(a + h)$ and expand.

f(a + h) = 2(a + h)^2 - (a + h) = 2a^2 + 4ah + 2h^2 - a - h

Step 2: subtract $f(a) = 2a^2 - a$ in parentheses.

(2a^2 + 4ah + 2h^2 - a - h) - (2a^2 - a) = 4ah + 2h^2 - h

Step 3: factor out $h$ and cancel, allowed because $h \neq 0$ .

\frac{4ah + 2h^2 - h}{h} = \frac{h(4a + 2h - 1)}{h} = 4a + 2h - 1

Check with $a = 1$ , $h = 1$ : $\dfrac{f(2) - f(1)}{1} = 6 - 1 = 5$ and $4(1) + 2(1) - 1 = 5$ ✓

Problem 4 · Does the Equation Define a Function?

Which equation does not define $y$ as a function of $x$ ?

✅ Correct! The input

x = 4

gives

y^2 = 4

, so

y = 2

y = -2

: one input, two outputs, as on the circle.

❌ This one is a function. Solving for

y

gives

y = 3 - \tfrac{3}{4}x

: one output per input.

❌ This one is a function. As

y

increases,

y^3 + y

increases, so each

x

comes from exactly one

y

. Like

x = y + 2^y

, the rule is implicit.

❌ This one is a function:

y = x^2 + 4

. The inputs

1

and

-1

share the output

5

, which a function allows; what it forbids is one input with two outputs.

❌ Not quite. Look for an input that gives two outputs.

Show solution

Try to solve each equation for $y$ :

$3x + 4y = 12 \Rightarrow y = 3 - \tfrac{3}{4}x$ : one output per input.
$x = y^2 \Rightarrow y = \pm\sqrt{x}$ : for $x = 4$ , both $y = 2$ and $y = -2$ work. Not a function.
$x = y^3 + y$ : no simple formula, but $y^3 + y$ increases with $y$ , so each $x$ has exactly one $y$ (an implicit function).
$y - x^2 = 4 \Rightarrow y = x^2 + 4$ : one output per input.

Solved: 0 / 4