Mathematics • Year 9 • Unit 2 • Lesson 9

Intercepts, Mixed Challenge

Mix every intercept skill: solve, predict, reverse-engineer. Catch one classic mistake, then invent four parabolas with prescribed intercept patterns.

Master · Mixed Challenge

1. Mixed problems

Each problem mixes two or more intercept skills. Show working. 3 marks each

1.1 Find the $y$-intercept AND $x$-intercepts of: (a) $y = x^2 - 36$ (b) $y = (x + 3)^2 - 4$ (c) $y = 3x^2 - 12$.

1.2 Without solving algebraically, state how many $x$-intercepts each parabola has. Justify using vertex and direction. (a) $y = (x - 4)^2 + 1$ (b) $y = -(x + 2)^2$ (c) $y = 2(x - 1)^2 - 8$ (d) $y = -3(x + 5)^2 - 4$.

1.3 A parabola in the form $y = (x - h)^2 + k$ has $x$-intercepts at $(1, 0)$ and $(7, 0)$. (a) State the axis of symmetry using the midpoint of the two roots. (b) Hence state $h$. (c) Find $k$ by substituting one intercept and solving. (d) Write the equation and find the $y$-intercept.

1.4 Find the $x$-intercepts of $y = 2(x - 1)^2 - 8$. Then check by computing the $y$-value at each $x$-intercept.

1.5 For each statement, decide TRUE or FALSE. If false, write the corrected statement. (a) "$y = x^2 + 4$ has $x$-intercepts at $x = \pm 2$." (b) "Every parabola has at least one $x$-intercept." (c) "Every parabola has exactly one $y$-intercept." (d) "$(x - 1)^2 = 9 \Rightarrow x - 1 = 3$, so $x = 4$."

1.6 A parabola has vertex $(2, -9)$ and one $x$-intercept at $(5, 0)$. (a) Use symmetry to find the OTHER $x$-intercept. (b) Use the vertex form $y = a(x - 2)^2 - 9$ with the point $(5, 0)$ to find $a$. (c) Write the full equation.

Stuck on 1.6(a)? Axis $x = 2$; intercept at $x = 5$ is $3$ units right of axis $\Rightarrow$ other intercept is $3$ units LEFT of axis, at $x = -1$.

2. Find the mistake

A classmate has found intercepts for five parabolas. Exactly two are wrong. Spot them, explain why, and fix them. 3 marks

Student's answers:

A: $y$-int of $y = x^2 - 4$ is $(0, -4)$. ✓

B: $x$-ints of $y = x^2 - 4$ are at $x = \pm 4$. (Take $\pm$ of the constant.)

C: $x$-int of $y = (x + 3)^2$ is $x = -3$ (repeated). ✓

D: $(x - 1)^2 = 16 \Rightarrow x - 1 = 4 \Rightarrow x = 5$, so one $x$-intercept at $5$.

E: $y = -2x^2 - 5$ has 0 $x$-intercepts (vertex $(0, -5)$ below; opens DOWN $\Rightarrow$ same side, no crossings). ✓

(a) Which two are wrong?

(b) For each wrong one, explain in one sentence why the student's reasoning is mistaken.

(c) Write the corrected answer for each wrong one.

Stuck on B? The constant term $-4$ is the $y$-intercept, not the $x$-intercept. For $x$-intercepts, you have to SET $y = 0$ and SOLVE.

3. Open-ended challenge, design parabolas by intercept count

Many valid answers. Be creative but precise. 4 marks

3.1 Invent FOUR parabolas in vertex form $y = a(x - h)^2 + k$, one matching each description:

A. TWO $x$-intercepts, opens UP.
B. TWO $x$-intercepts, opens DOWN.
C. EXACTLY ONE $x$-intercept (vertex on the $x$-axis).
D. NO $x$-intercepts at all.

For each:
(i) Write the equation.
(ii) State the vertex.
(iii) Find the $y$-intercept.
(iv) Confirm the intercept count by checking signs of $a$ and $k$ (or showing the algebra).

Bonus: Use a DIFFERENT non-zero $h$ for each of the four (i.e. none have vertex on the $y$-axis).

Stuck? For A: $a$ and $k$ opposite signs (e.g. $a > 0$, $k < 0$). For C: $k = 0$. For D: $a$ and $k$ same signs.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

1.1, Both intercepts

(a) $y = x^2 - 36$: $y$-int $(0, -36)$. $x^2 = 36 \Rightarrow x = \pm 6$, so $x$-ints $(6, 0)$ and $(-6, 0)$.
(b) $y = (x + 3)^2 - 4$: $y$-int $(0, 9 - 4) = (0, 5)$. $(x + 3)^2 = 4 \Rightarrow x + 3 = \pm 2 \Rightarrow x = -1$ or $x = -5$, so $x$-ints $(-1, 0)$ and $(-5, 0)$.
(c) $y = 3x^2 - 12$: $y$-int $(0, -12)$. $3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$, so $x$-ints $(2, 0)$ and $(-2, 0)$.

1.2, Predict the count

(a) Vertex $(4, 1)$ above; opens up. Same side $\Rightarrow$ 0 $x$-ints.
(b) Vertex $(-2, 0)$ ON the $x$-axis $\Rightarrow$ 1 $x$-int (repeated at $x = -2$).
(c) Vertex $(1, -8)$ below; opens up. Opposite sides $\Rightarrow$ 2 $x$-ints.
(d) Vertex $(-5, -4)$ below; opens down. Same side $\Rightarrow$ 0 $x$-ints.

1.3, Reverse from intercepts $(1, 0)$ and $(7, 0)$

(a) Midpoint of $1$ and $7$: $(1 + 7)/2 = 4$. Axis $x = 4$. (b) $h = 4$. (c) Sub $(1, 0)$ into $y = (x - 4)^2 + k$: $0 = (1 - 4)^2 + k = 9 + k \Rightarrow k = -9$. (d) Equation: $y = (x - 4)^2 - 9$. $y$-intercept: $y = (0 - 4)^2 - 9 = 16 - 9 = 7$, so $(0, 7)$.

1.4, $y = 2(x - 1)^2 - 8$

$0 = 2(x - 1)^2 - 8 \Rightarrow (x - 1)^2 = 4 \Rightarrow x - 1 = \pm 2 \Rightarrow x = 3$ or $x = -1$. Check $x = 3$: $y = 2(2)^2 - 8 = 8 - 8 = 0$ ✓. Check $x = -1$: $y = 2(-2)^2 - 8 = 8 - 8 = 0$ ✓. $x$-intercepts $(3, 0)$ and $(-1, 0)$.

1.5, True / False

(a) FALSE. $x^2 + 4 = 0 \Rightarrow x^2 = -4$, no real solutions. Zero $x$-intercepts.
(b) FALSE. A parabola can have 0, 1, or 2 $x$-intercepts. Some never cross the $x$-axis.
(c) TRUE. Sub $x = 0$ always gives exactly one $y$-value.
(d) FALSE. Missing the $\pm$. $(x - 1)^2 = 9 \Rightarrow x - 1 = \pm 3 \Rightarrow x = 4$ or $x = -2$.

1.6, Vertex $(2, -9)$, $x$-int $(5, 0)$

(a) Axis $x = 2$; $(5, 0)$ is $3$ right of axis, so other $x$-int is $3$ left of axis at $x = -1$. So $(-1, 0)$. (b) Sub $(5, 0)$: $0 = a(5 - 2)^2 - 9 = 9a - 9 \Rightarrow 9a = 9 \Rightarrow a = 1$. (c) Equation: $y = (x - 2)^2 - 9$.

2, Find the mistake

(a) The two wrong answers are B and D.
(b) B: The student treated the constant $-4$ as if it were the $x$-intercept (it's actually the $y$-intercept). To find $x$-intercepts, you have to set $y = 0$ and SOLVE for $x$. D: The student forgot the $\pm$ when taking the square root, missing half the solutions.
(c) B corrected: $0 = x^2 - 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. $x$-intercepts at $(2, 0)$ and $(-2, 0)$. D corrected: $(x - 1)^2 = 16 \Rightarrow x - 1 = \pm 4 \Rightarrow x = 5$ or $x = -3$. Two $x$-intercepts.
Both are flagged in the lesson's "Common Pitfalls": reading the wrong intercept off the equation, and square-rooting one side only.

3, Open-ended challenge (sample solutions)

A. (2 $x$-ints, opens up): $y = (x - 2)^2 - 9$. Vertex $(2, -9)$. $y$-int: $(0 - 2)^2 - 9 = -5$, so $(0, -5)$. Check: $a = 1 > 0$, $k = -9 < 0$, opposite signs $\Rightarrow$ 2 $x$-ints. ✓

B. (2 $x$-ints, opens down): $y = -(x + 1)^2 + 4$. Vertex $(-1, 4)$. $y$-int: $-(1)^2 + 4 = 3$, so $(0, 3)$. Check: $a = -1 < 0$, $k = 4 > 0$, opposite signs $\Rightarrow$ 2 $x$-ints. ✓

C. (1 $x$-int, vertex on axis): $y = 2(x - 3)^2$. Vertex $(3, 0)$. $y$-int: $2(9) = 18$, so $(0, 18)$. Check: $k = 0$ $\Rightarrow$ exactly 1 (repeated) $x$-int at $x = 3$. ✓

D. (0 $x$-ints): $y = (x + 4)^2 + 5$. Vertex $(-4, 5)$. $y$-int: $(4)^2 + 5 = 21$, so $(0, 21)$. Check: $a = 1 > 0$, $k = 5 > 0$, same signs $\Rightarrow$ 0 $x$-ints. ✓

Marking: 1 mark per parabola with valid equation, vertex, $y$-intercept and correct intercept-count justification. Full marks for four parabolas with four DIFFERENT $h$-values.