Rewrite the quadratic expression $-3x^2 + 18x - 14$ in the vertex form $a(x - h)^2 + k$ by completing the square, and determine the exact coordinate pair $(h, k)$ corresponding to the maximum value of the function.

Final answer

$(h, k) = (3, 13)$ , with equivalent vertex form $-3(x - 3)^2 + 13$

Steps

Factor out the leading coefficient from the variable terms

To rewrite the quadratic expression $-3x^2 + 18x - 14$ in vertex form, $a(x - h)^2 + k$ , we use the method of completing the square.

Factor the coefficient of $x^2$ , which is $-3$ , from the first two terms: $-3x^2 + 18x - 14 = -3(x^2 - 6x) - 14$

Complete the square inside the parentheses

To turn $x^2 - 6x$ into a perfect square trinomial of the form $(x - d)^2 = x^2 - 2dx + d^2$ :

Take half of the coefficient of $x$ : $\frac{-6}{2} = -3$
Square this value: $(-3)^2 = 9$

Add and subtract $9$ inside the parentheses so the overall value does not change: $-3(x^2 - 6x + 9 - 9) - 14$

Write the trinomial as a squared binomial: $-3\bigl((x - 3)^2 - 9\bigr) - 14$

Common Pitfall: When removing the subtracted constant ( $-9$ ) from the parentheses, remember that it is multiplied by the outer factor of $-3$ . Forgetting to multiply by $-3$ is a frequent source of sign errors.

Distribute and combine constants

Distribute the factor $-3$ : $-3(x - 3)^2 + (-3)(-9) - 14$ $-3(x - 3)^2 + 27 - 14$ $-3(x - 3)^2 + 13$

The expression in vertex form is: $-3(x - 3)^2 + 13$

Identify the vertex $(h, k)$

Comparing $-3(x - 3)^2 + 13$ to the standard vertex form $a(x - h)^2 + k$ :

$a = -3$
$h = 3$
$k = 13$

Because $a = -3 < 0$ , the parabola opens downward, which means the vertex represents the maximum value of the function. Therefore, the coordinate pair corresponding to the maximum is: $(h, k) = (3, 13)$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to multiply the added squared term inside the parentheses by the leading coefficient -3 when balancing the constant outside.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$-3x^2 + 18x - 14 = -3(x^2 - 6x) - 14$
Step 2
$= -3(x^2 - 6x + 9) - 14 - 9$
Correct
$-3(x^2 - 6x + 9) - 14 - (-3)(9) = -3(x^2 - 6x + 9) - 14 + 27$
Adding $9$ inside parentheses multiplied by $-3$ changes the overall value by $-3 \times 9 = -27$ . To balance the expression, you must add $27$ (or subtract $-27$ ) outside, rather than subtracting $9$ .
Step 3
$= -3(x - 3)^2 - 23$
Correct
$-3(x - 3)^2 + 13$
The binomial $(x - 3)^2$ was formed correctly from $x^2 - 6x + 9$ , but the constant term $-23$ inherits the balancing error from the previous step.
Step 4
$(h, k) = (3, -23)$
Correct
$(h, k) = (3, 13)$
The $x$ -coordinate $h = 3$ is correct, but the maximum value $k$ is incorrect due to the constant term error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly factored the leading coefficient $-3$ from both variable terms.

You found the correct value $(\frac{-6}{2})^2 = 9$ needed to complete the square.

You correctly recognized that $h = 3$ from the binomial factor $(x - 3)^2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributive Balancing Error

You balanced the $+9$ inside the parentheses by subtracting $9$ outside, forgetting that the inner $+9$ is multiplied by the leading factor $-3$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating the constant added inside parentheses as having a coefficient of $1$ rather than distributing the outer coefficient $a$ when keeping the expression equivalent.

REASONING PATH

Your ACEpath teacher’s suggested approach

Account for the leading coefficient when balancing: In $a(x^2 + bx + c) + d$ , adding $c$ inside the parentheses actually adds $a \cdot c$ to the entire expression.
Always balance an expression by subtracting $a \cdot c$ outside, so that $a(x^2 + bx + c) - a \cdot c$ preserves equality.

CORRECTED PATH

ACEpath Teacher's Demonstration

Write the balancing term explicitly: $-3(x^2 - 6x + 9) - 14 - (-3)(9)$
Simplify the outside constants: $-14 + 27 = 13$
Express in vertex form: $-3(x - 3)^2 + 13$
Read the vertex $(h, k)$ directly from $a(x - h)^2 + k$ : $(h, k) = (3, 13)$

Final answer

$(h, k) = (3, 13)$ , with vertex form $-3(x - 3)^2 + 13$

NEXT STEPS

Final Takeaways from Your Teacher

Practice multiplying the inside constant by the factored coefficient before writing the balancing term outside.
Check your vertex form by expanding it back out to verify that it matches the original expression.

OverviewACEpath Teacher's Diagnosis

02 The student fails to factor the negative sign out of the linear term, writing + 6x instead of - 6x inside the parentheses.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
-3x^2 + 18x - 14 = -3(x^2 + 6x) - 14
Correct
-3(x^2 - 6x) - 14
When factoring out $-3$ from the linear term $+18x$ , dividing $+18$ by $-3$ yields $-6x$ , not $+6x$ .
Step 2
-3(x^2 + 6x + 9) - 14 + 27 $\implies$ -3(x + 3)^2 + 13
Correct
-3(x^2 - 6x + 9) - 14 + 27 $\implies$ -3(x - 3)^2 + 13
Your technique for completing the square and balancing the subtracted value with $+27$ was algebraically consistent with your previous step, but the binomial inside is affected by the earlier sign error.
Step 3
(h, k) = (-3, 13)
Correct
(h, k) = (3, 13)
You accurately extracted the vertex $(h, k)$ from your expression $-3(x + 3)^2 + 13$ , but the correct vertex form $-3(x - 3)^2 + 13$ gives $(h, k) = (3, 13)$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You recognized the need to factor the leading coefficient out of the variable terms before completing the square.

You correctly calculated the square of half the linear coefficient to form a perfect square trinomial.

You correctly accounted for the outer multiplier when balancing the constant term on the outside.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Factoring negative leading coefficient sign error

You wrote $+6x$ instead of $-6x$ inside the parentheses after factoring out $-3$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the sign inversion that occurs when dividing a positive term by a negative factor.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distributive Check: Factoring is division; multiplying back must reproduce the original expression exactly.
Mentally expand $-3(x^2 \pm bx)$ right away to verify that the signs match the original terms before completing the square.

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor $-3$ correctly from both variable terms: $-3(x^2 - 6x) - 14$ .
Complete the square with $\left(\frac{-6}{2}\right)^2 = 9$ : $-3(x^2 - 6x + 9) - 14 - (-3)(9) = -3(x - 3)^2 - 14 + 27$ .
Combine constant terms: $-3(x - 3)^2 + 13$ .

Final answer

(h, k) = (3, 13)

NEXT STEPS

Final Takeaways from Your Teacher

Practice factoring negative coefficients from binomials with mixed signs.
Always perform a quick mental distribution step to double-check signs before adding the square term.

OverviewACEpath Teacher's Diagnosis

03 The student misinterprets the sign of h in vertex form a(x - h)² + k, incorrectly taking h as -3 instead of 3.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
=-3(x^2 - 6x) - 14
Step 2
=-3(x^2 - 6x + 9) - 14 + 27 ⟹ -3(x - 3)^2 + 13
Step 3
(h, k) = (-3, 13)
Correct
(h, k) = (3, 13)
In vertex form $a(x - h)^2 + k$ , the expression inside is $(x - h)$ . Comparing $(x - h)$ to $(x - 3)$ gives $h = 3$ , not $-3$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly factored out the leading coefficient from the quadratic and linear terms.

You accurately completed the square and balanced the constant term across parentheses.

You correctly determined the vertical coordinate $k = 13$ of the vertex.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Vertex Sign Reversal

You read $(x - 3)$ as $h = -3$ instead of recognizing that $x - h = x - 3$ means $h = 3$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the sign in the factored binomial $(x - h)$ with the coordinate value $h$ , forgetting that the vertex occurs where the squared term vanishes ( $x - 3 = 0 \implies x = 3$ ).

REASONING PATH

Your ACEpath teacher’s suggested approach

Finding h via Zero of the Square: The horizontal coordinate $h$ always satisfies $x - h = 0$ , so $x = h$ .
Set the inside of the squared parenthesis to zero and solve for $x$ to confirm the sign of $h$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Compare $-3(x - 3)^2 + 13$ with the vertex form $a(x - h)^2 + k$ .
Match terms: $a = -3$ , $x - h = x - 3 \implies h = 3$ , and $k = 13$ .

Final answer

(h, k) = (3, 13)

NEXT STEPS

Final Takeaways from Your Teacher

Whenever you read a vertex from $a(x - h)^2 + k$ , solve $(x - h) = 0$ as a quick check for the sign of the $x$ -coordinate.
Practice extracting vertices from forms with both $(x - h)$ and $(x + h)$ to make the sign switch automatic.

Lock it in

Question

When completing the square in $a(x^2 + bx) + c$ , what value must be added outside to balance adding $d$ inside the parentheses?

Answer

Subtract $a \cdot d$ outside. Because the $d$ inside is scaled by $a$ , the net change is $a \cdot d$ , which requires $-a \cdot d$ to balance.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to multiply the added squared term inside the parentheses by the leading coefficient -3 when balancing the constant outside.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student fails to factor the negative sign out of the linear term, writing + 6x instead of - 6x inside the parentheses.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student misinterprets the sign of h in vertex form a(x - h)² + k, incorrectly taking h as -3 instead of 3.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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