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Quadratic Functions and Parabolas: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A rewritten rule

Classify $f(x)=(x+2)^2-x(x+4)+7$ , once simplified, as quadratic, linear, or constant.
1. Hint 1
  The visible square does not decide the degree before like terms are combined.
2. Hint 2
  Expand both products and compare their variable terms.
Answer

Constant; $f(x)=11$ .

Full solution

$\begin{aligned} f(x)&=x^2+4x+4-x^2-4x+7 \\ f(x)&=11 \end{aligned}$

The squared and linear terms cancel, so the leading quadratic coefficient is zero.
Its graph is the horizontal line $y=11$ , not a parabola.

Answer

Constant; $f(x)=11$ .

Key idea

A quadratic needs a nonzero squared term after simplification.
Problem 1: Mark complete
Problem 2 Where a rule begins

Find the $y$ -intercept of $g(x)=4-x(3x+2)$ .
1. Hint 1
  An intercept on the vertical axis has input zero.
2. Hint 2
  Evaluate the rule before expanding if that is shorter.
Answer

$(0,4)$ .

Full solution

The $y$ -axis has $x=0$ .
$g(0)=4-0(3(0)+2)$
This gives $g(0)=4$ , so the intercept is $(0,4)$ .

Answer

$(0,4)$ .

Key idea

The vertical intercept is the output at input zero.
Problem 2: Mark complete
Problem 3 A parent point moves

The parent parabola $y=x^2$ is reflected across the $x$ -axis, stretched vertically by 3, then shifted 2 units right and 1 unit down. Where does its point $(-1,1)$ move?
1. Hint 1
  Track the horizontal coordinate separately from the output changes.
2. Hint 2
  Apply the listed output operations in order, then the horizontal shift.
Answer

$(1,-4)$ .

Full solution

The input moves from $-1$ to $-1+2=1$ .
The output is reflected and scaled, then lowered.
$y=-3(1)-1$
Hence the image is $(1,-4)$ .
It checks in the transformed rule
$y=-3(x-2)^2-1.$

Answer

$(1,-4)$ .

Key idea

Horizontal and output transformations act on different coordinates of a graph point.
Problem 3: Mark complete
Problem 4 A curve on the grid

The figure shows a quadratic function $f$ . Write $f$ in vertex and standard form, and state its range.

The graph, with two of its points marked.

Text description of this figure

A grid with the horizontal axis running from negative 5 to 3 and the vertical axis running from negative 10 to 2, gridlines and number labels at every whole number. A single smooth curve enters through the top edge of the grid a little to the left of x equals negative 4, descends to a marked low point at negative 1 comma negative 9, then rises back up and leaves through the top edge a little to the right of x equals 2. A second marked point sits on the rising right side of the curve at 1 comma negative 5.
1. Hint 1
  The turning point gives two entries of vertex form.
2. Hint 2
  Use the other marked point to determine the multiplier of the square.
Answer

Vertex form $(x+1)^2-9$ ; standard form $x^2+2x-8$ ; range $[-9,\infty)$ .

Full solution

The vertex is $(-1,-9)$ , so write
$f(x)=a(x+1)^2-9.$
The marked point $(1,-5)$ gives $-5=4a-9$ , hence $a=1$ .
Expanding gives
$f(x)=x^2+2x-8$
Since $a>0$ , the minimum is $-9$ , so the range is $[-9,\infty)$ .

Answer

Vertex form $(x+1)^2-9$ ; standard form $x^2+2x-8$ ; range $[-9,\infty)$ .

Key idea

A vertex and one other point determine the coefficient needed to recover a quadratic rule.
Problem 4: Mark complete
Problem 5 Moving a minimum

The function $f(x)=2(x-1)^2+3$ is changed to $g(x)=f(x+4)-5$ . Find the vertex, opening direction, and range of $g$ .
1. Hint 1
  Substitute the whole new input into the original rule.
2. Hint 2
  Combine the horizontal and vertical shifts before reading the vertex.
Answer

Vertex $(-3,-2)$ ; opens upward; range $[-2,\infty)$ .

Full solution

Substitute $x+4$ for the original input.
$\begin{aligned} g(x)&=2(x+4-1)^2+3-5 \\ g(x)&=2(x+3)^2-2 \end{aligned}$

The vertex is $(-3,-2)$ .
The positive multiplier makes it the minimum; all outputs are at least $-2$ .

Answer

Vertex $(-3,-2)$ ; opens upward; range $[-2,\infty)$ .

Key idea

Substituting a shifted input into vertex form reveals the combined movement.
Problem 5: Mark complete
Problem 6 Two labels for one rule

A quadratic is given in factored form as $(x+4)(x-10)$ . Find its standard form $ax^2+bx+c$ and its vertex form $(x-h)^2+k$ .
1. Hint 1
  Expand the factored form to reach standard form.
2. Hint 2
  The axis sits halfway between the zeros; evaluate there for the vertex height.
Answer

$x^2-6x-40$ ; vertex form $(x-3)^2-49$ .

Full solution

Expand the factored form.
$(x+4)(x-10)=x^2-6x-40$

The zeros are $-4$ and $10$ , so the axis is their midpoint, $x=3$ .
Evaluate the factored form there.
$\begin{aligned} f(3)&=(3+4)(3-10) \\ f(3)&=7(-7) \\ f(3)&=-49 \end{aligned}$
So the vertex form is $(x-3)^2-49$ .

Answer

$x^2-6x-40$ ; vertex form $(x-3)^2-49$ .

Key idea

The axis is the midpoint of the zeros, and evaluating there gives the vertex height directly from factored form.
Problem 6: Mark complete
Problem 7 Reading the intercept and the ends

For $f(x)=2x^2-3x-6$ , give the $y$ -intercept and explain the end behavior as $x$ grows large positively or negatively.
1. Hint 1
  The constant term sets the output at zero.
2. Hint 2
  For large nonzero inputs, factor out the highest power.
Answer

$y$ -intercept $(0,-6)$ ; both ends rise without bound.

Full solution

The intercept is $f(0)=-6$ .
For $x\ne0$ ,
$f(x)=x^2\left(2-\tfrac{3}{x}-\tfrac{6}{x^2}\right)$
The bracket approaches positive $2$ as the magnitude of $x$ grows.
The squared factor grows positively in both directions, so the outputs rise without bound at both ends.

Answer

$y$ -intercept $(0,-6)$ ; both ends rise without bound.

Key idea

The leading coefficient controls distant behavior even when the vertical intercept is negative.
Problem 7: Mark complete
Problem 8 Two equal readings

A quadratic satisfies $f(-2)=f(6)=9$ . A student locates its axis at $x=4$ by taking half the distance between the inputs. Is that correct? Explain and locate the axis.
1. Hint 1
  A distance and a coordinate are different quantities.
2. Hint 2
  Locate the midpoint of the two equal-height inputs.
Answer

No; the axis is $x=2$ .

Full solution

The distance between the inputs is $8$ , and half that distance is $4$ .
But starting at $-2$ and moving $4$ reaches $2$ .
Equivalently,
$h=\frac{-2+6}{2}$
Therefore $h=2$ .
Equal-height points on a parabola are reflected across this midpoint, not across half their distance measured from zero.

Answer

No; the axis is $x=2$ .

Key idea

A symmetry axis is a midpoint coordinate, not half a separation.
Problem 8: Mark complete
Problem 9 Opposite inputs

For $f(x)=ax^2+bx+c$ with $a\ne0$ , a student expands $f(h+1)-f(h-1)$ and gets $4ah+2b$ . They say a vertical symmetry axis $x=h$ must therefore satisfy $h=-b/(2a)$ . Is this reasoning valid? Explain.
1. Hint 1
  Points one unit to either side of a mirror must have equal heights.
2. Hint 2
  Set their height difference equal to zero and check the divisor.
Answer

Yes; any vertical symmetry axis must have $h=-\frac{b}{2a}$ .

Full solution

The squared terms differ by $4h$ and the linear inputs differ by $2$ , so the stated expansion is correct.
Symmetry requires
$\begin{aligned} 4ah+2b&=0 \\ 2ah+b&=0 \\ h&=-\frac{b}{2a} \end{aligned}$

Division is legal because $a\ne0$ .
This proves where a vertical symmetry axis must lie, consistently with the known parabola symmetry.

Answer

Yes; any vertical symmetry axis must have $h=-\frac{b}{2a}$ .

Key idea

Equal heights on opposite sides of an axis force a relation among the quadratic’s coefficients.
Problem 9: Mark complete
Problem 10 A shared pair of zeros

Two quadratics have zeros $-3$ and $5$ . One has leading coefficient $1$ and the other has leading coefficient $-2$ . Are their vertices identical? Explain and give both vertices.
1. Hint 1
  The midpoint of the zeros fixes the horizontal coordinate for both.
2. Hint 2
  Evaluate each factored rule at that midpoint to compare heights.
Answer

No; the vertices are $(1,-16)$ and $(1,32)$ , respectively.

Full solution

The common axis is the midpoint of the zeros.
$\begin{aligned} x&=\frac{-3+5}{2} \\ x&=1 \end{aligned}$
The first rule is $(x+3)(x-5)$ and the second is $-2(x+3)(x-5)$ .
At the axis,
$\begin{aligned} f(1)&=4(-4) \\ f(1)&=-16 \\ g(1)&=-2(4)(-4) \\ g(1)&=32 \end{aligned}$

The leading coefficient changes the vertex height and opening while the axis stays fixed.

Answer

No; the vertices are $(1,-16)$ and $(1,32)$ , respectively.

Key idea

Two zeros determine a common axis but leave the parabola’s vertical scale free.
Problem 10: Mark complete

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