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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 missing coefficient

The graph of $y=3x^2+bx+2$ has axis of symmetry $x=-2$ . Find $b$ .
1. Hint 1
  The axis location depends on the leading and linear coefficients.
2. Hint 2
  Substitute the known axis coordinate into $-b/(2a)$ .
Answer

$b=12$ .

Full solution

Here $a=3$ , so the given axis requires
$-\frac b6=-2$

Multiplying by $-6$ gives $b=12$ .
Checking the axis of $3x^2+12x+2$ gives $-12/6=-2$ .

Answer

$b=12$ .

Key idea

An axis of symmetry can determine a missing linear coefficient.
Problem 1: Mark complete
Problem 2 A marked curve

The figure shows a parabola with equation $y=a(x+1)^2+2$ . Find $a$ from the marked point $P$ .

A parabola with one marked point $P$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative five to three and the vertical y-axis from negative three to three, with gridlines, tick marks and number labels at every whole number, the origin labeled 0, and arrowheads at both ends of each axis. A smooth downward opening parabola is drawn: it rises from the bottom edge of the grid on the left, turns at its highest point, which sits one unit left of the y-axis and two units above the x-axis, and falls again to the bottom edge on the right. One point of the curve carries a filled dot labeled with the letter P only; that dot sits on the x-axis, one unit to the right of the origin. No coordinates, equation, vertex marker, axis of symmetry or coefficient is shown.
1. Hint 1
  The coordinates of a point on the curve satisfy its equation.
2. Hint 2
  Read the marked point from the grid and substitute its two coordinates.
Answer

$a=-\dfrac12$ , or $-0.5$ .

Full solution

The marked point is $P=(1,0)$ .
Substitute it.
$0=a(1+1)^2+2$

Thus $4a=-2$ , giving $a=-1/2$ .
A negative coefficient agrees with the downward opening in the figure.

Answer

$a=-\dfrac12$ , or $-0.5$ .

Key idea

A point away from a known vertex determines the leading coefficient of a parabola.
Problem 2: Mark complete
Problem 3 A vertex condition

The highest point of $y=-2(x-3)^2+k$ also lies on the line $y=2x+1$ . Find $k$ .
1. Hint 1
  The squared term identifies the input at the highest point.
2. Hint 2
  Use that input in the line equation to find the shared height.
Answer

$k=7$ .

Full solution

The negative coefficient makes the vertex the highest point.
Its coordinates are $(3,k)$ .
Because that vertex lies on the line, its height must equal the line's height at $x=3$ .
$k=2(3)+1$

Therefore $k=7$ .
At $x=3$ , the squared term vanishes and both equations give $7$ .

Answer

$k=7$ .

Key idea

A condition on the vertex can recover the constant outside a squared binomial.
Problem 3: Mark complete
Problem 4 A product graph

For $y=(x-3)(3x+3)+4$ , give the vertex form, vertex, axis of symmetry, opening direction, and y-intercept. Also state the number of x-intercepts.
1. Hint 1
  Expand the product before identifying the quadratic coefficients.
2. Hint 2
  Complete the square to locate the vertex and axis, and use the leading coefficient for the direction.
3. Hint 3
  Set $x=0$ for the y-intercept and inspect the discriminant for the x-intercept count.
Answer

$y=3(x-1)^2-8$ ; vertex $(1,-8)$ ; axis $x=1$ ; opens upward; y-intercept $(0,-5)$ ; two x-intercepts.

Full solution

Expansion gives
$y=3x^2-6x-5$

Factor $3$ from the two $x$ terms.
$y=3(x^2-2x)-5$

Half of $-2$ is $-1$ , and $(-1)^2=1$ , so
$x^2-2x=(x-1)^2-1.$
Multiplying that $-1$ by $3$ and combining with $-5$ gives $-8$ .
$y=3(x-1)^2-8$

The vertex is $(1,-8)$ and the axis is $x=1$ .
Since $a=3>0$ , the curve opens upward.
At $x=0$ , $y=-5$ , so the y-intercept is $(0,-5)$ .

The discriminant is
$D=(-6)^2-4(3)(-5)$

This is $96>0$ , so there are two x-intercepts, consistent with an upward curve whose lowest point lies below the x-axis.

Answer

$y=3(x-1)^2-8$ ; vertex $(1,-8)$ ; axis $x=1$ ; opens upward; y-intercept $(0,-5)$ ; two x-intercepts.

Key idea

Expanding and completing the square expose different features of the same parabola.
Problem 4: Mark complete
Problem 5 An unfinished graph

The figure shows the part of $y=\tfrac12x^2+2x-2$ with $x\le-4$ . Complete the graph for $x>-4$ on the same grid, and label the vertex and y-intercept.

The drawn part of the graph, for $x \le -4$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative seven to three and the vertical y-axis from negative five to five, with gridlines, tick marks and number labels at every whole number, the origin labeled 0, and arrowheads at both ends of each axis. Only part of a curve is drawn, and all of it lies to the left of the vertical line four units left of the origin. It enters the grid at the top edge, a little past six units to the left of the origin, falls steadily to the right, and stops at the point four units left of the origin and two units below the x-axis. The rest of the grid is blank: nothing is drawn to the right of that end point, and there is no marked vertex, no marked intercept, no axis of symmetry line, no coordinates and no equation.
1. Hint 1
  The two sides of a parabola are mirror images of each other across its axis of symmetry.
2. Hint 2
  Find the axis from the coefficients, then use equal horizontal distances to place matching points.
Answer

Vertex $(-2,-4)$ ; y-intercept $(0,-2)$ ; the right half mirrors the left half across $x=-2$ .

Full solution

The axis is
$x=-\frac{2}{2(1/2)}$

This is $x=-2$ .
Substitution gives vertex height $-4$ .
Equivalently,
$y=\frac12(x+2)^2-4$

Plot the vertex, then reflect each shown point across $x=-2$ .
For example, the drawn end point $(-4,-2)$ matches $(0,-2)$ , and $(-6,4)$ matches $(2,4)$ .
Draw a smooth curve from $(-4,-2)$ down through the vertex and up through $(0,-2)$ and $(2,4)$ .
At $x=0$ , the equation gives $-2$ , checking the y-intercept.

The completed graph, with the vertex and y-intercept labeled.

Answer

Vertex $(-2,-4)$ ; y-intercept $(0,-2)$ ; the right half mirrors the left half across $x=-2$ .

Key idea

Once the axis is known, the vertex and mirrored points together complete a partly drawn parabola.
Problem 5: Mark complete
Problem 6 Two horizontal levels

Consider the parabola $y=-(x+1)^2+4$ . Find the horizontal distance between its two points at height $y=0$ , and the horizontal distance between its two points at height $y=3$ . Explain which level gives the wider span.
1. Hint 1
  At each given height, solve for the two x-coordinates.
2. Hint 2
  A horizontal distance is the larger x-coordinate minus the smaller.
Answer

At $y=0$ : $4$ units. At $y=3$ : $2$ units. The span at $y=0$ is wider.

Full solution

At height zero, $(x+1)^2=4$ , so the x-coordinates are $-3$ and $1$ .
Their horizontal separation is
$1-(-3)=4$

At height three, $(x+1)^2=1$ , so the x-coordinates are $-2$ and $0$ .
Their separation is
$0-(-2)=2$

The lower level has the wider span because it lies farther below this downward parabola’s highest point.

Answer

At $y=0$ : $4$ units. At $y=3$ : $2$ units. The span at $y=0$ is wider.

Key idea

Horizontal spans of a parabola come from the separation of equal-height inputs.
Problem 6: Mark complete
Problem 7 Two constant terms

Compare $y=2x^2-16x+35$ with $y=2x^2-16x+30$ . Find the axis of symmetry and the number of x-intercepts of each graph.
1. Hint 1
  An axis depends on $a$ and $b$ , which are unchanged.
2. Hint 2
  Compute each discriminant with its own signed constant term.
Answer

Common axis $x=4$ . First graph: no x-intercepts. Second graph: two x-intercepts.

Full solution

For both graphs, the axis is
$x=-\frac{-16}{2(2)}$

Thus the shared axis is $x=4$ .
With $a=2$ and $b=-16$ , the first discriminant is $(-16)^2-4(2)(35)$ , which is $256-280=-24$ , so the first graph has no x-intercepts.
The second discriminant is $(-16)^2-4(2)(30)$ , which is $256-240=16$ , so the second graph has two.

Substituting $x=4$ gives vertex heights $3$ and $-2$ , respectively, which confirms the counts for upward parabolas.

Answer

Common axis $x=4$ . First graph: no x-intercepts. Second graph: two x-intercepts.

Key idea

A change in the constant can alter intercept counts without moving the axis of symmetry.
Problem 7: Mark complete
Problem 8 Matching graph features

A student says that two parabolas with the same width and the same y-intercept must have the same vertex. Is that correct? Justify your answer.
1. Hint 1
  The leading coefficient controls width, while the constant controls the y-intercept.
2. Hint 2
  Try keeping those coefficients fixed while changing the linear coefficient.
Answer

False; $y=x^2$ and $y=x^2-4x$ have vertices $(0,0)$ and $(2,-4)$ .

Full solution

Choose equations $y=x^2$ and $y=x^2-4x$ .
Both have leading coefficient $1$ and y-intercept $(0,0)$ , so their widths and y-intercepts match.

The second equation rewrites as
$y=(x-2)^2-4$

Its vertex is $(2,-4)$ , whereas the first vertex is $(0,0)$ .
These unequal vertices disprove the claim.

Answer

False; $y=x^2$ and $y=x^2-4x$ have vertices $(0,0)$ and $(2,-4)$ .

Key idea

Equal width and y-intercept do not determine the linear coefficient or the vertex.
Problem 8: Mark complete
Problem 9 A vertex on an axis

Let $y=ax^2+bx+c$ with real coefficients and $a\ne0$ . Explain what the vertex lying on the y-axis forces $b$ to be, and give the vertex's coordinates in that case.
1. Hint 1
  A point on the y-axis has x-coordinate zero.
2. Hint 2
  Use the formula for the vertex x-coordinate and the condition $a\ne0$ , then find the height at that input.
Answer

$b=0$ ; the vertex is $(0,c)$ , which is the y-intercept.

Full solution

The vertex x-coordinate must satisfy
$-\frac b{2a}=0$

Because $a\ne0$ , multiplication by $2a$ is valid and gives $-b=0$ .
Hence $b=0$ .
With that value, the axis is indeed $x=0$ .

Substituting $x=0$ into $y=ax^2+bx+c$ leaves $y=c$ .
So the vertex is $(0,c)$ , which is also the y-intercept.

Answer

$b=0$ ; the vertex is $(0,c)$ , which is the y-intercept.

Key idea

A quadratic vertex on the y-axis forces the linear coefficient to vanish, and the vertex is then the y-intercept.
Problem 9: Mark complete
Problem 10 A scaled equation

A quadratic $y=ax^2+bx+c$ has discriminant $D$ . Every coefficient is multiplied by $-2$ to make a new graph. A student says that the new graph has the same number of x-intercepts. Is the claim correct? Explain using the new discriminant.
1. Hint 1
  Substitute the scaled coefficients into the discriminant expression.
2. Hint 2
  Compare the sign of the new discriminant with the sign of the original.
Answer

Yes; the new discriminant is $4D$ .

Full solution

The new coefficients are $-2a$ , $-2b$ , and $-2c$ .
$D_{\text{new}}=4b^2-16ac$

Factor out $4$ .
$D_{\text{new}}=4D$

Multiplication by positive $4$ preserves whether the discriminant is positive, zero, or negative.
The number of x-intercepts therefore stays the same, although the opening direction reverses.

The same argument covers any nonzero multiplier $t$ .
Scaling every coefficient by $t$ gives
$D_{\text{new}}=t^2b^2-4t^2ac$

That expression equals $t^2D$ , and $t^2>0$ for every nonzero $t$ , so the sign of the discriminant never changes.

Answer

Yes; the new discriminant is $4D$ .

Key idea

Scaling every coefficient by a nonzero number preserves the number of real zeros.
Problem 10: Mark complete

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