Solutions: Polynomials - Test 2

Q1. (b) 3

The graph intersects the x-axis at 3 distinct points.

Q2. (c) A curve intersecting x-axis at 3 points

A quadratic polynomial (degree 2) can have at most 2 zeroes. A curve with 3 intersections represents a cubic or higher degree polynomial.

Q3. (c) $a < 0$

If the coefficient of $x^2$ is negative, the parabola opens downwards.

Q4. (a) Both A and R are true and R is the correct explanation of A.

$D = 0^2 - 4(1)(1) = -4 < 0$. No real roots, so graph does not touch x-axis.

Q5. The graph intersects the x-axis at $x = -2$ and $x = 2$.
Hence, the zeroes are -2 and 2. [2]

Q6. $p(x) = x^2 - 4$. Zeroes are $\pm 2$. Vertex at $(0, -4)$.

Parabola opening upwards cutting x-axis at -2, 2 and y-axis at -4. [2]

Q7. No. A quadratic polynomial can also touch the x-axis at exactly one point (equal roots). Example: $(x-1)^2$. So, it is not necessarily linear. [2]

Q8. Since it passes through $(-1, 0)$ and $(3, 0)$, zeroes are $\alpha = -1, \beta = 3$.
$p(x) = k(x - \alpha)(x - \beta) = k(x+1)(x-3) = k(x^2 - 2x - 3)$. [1]
It passes through $(0, -3)$. Put $x=0, y=-3$:
$-3 = k(0 - 0 - 3) \Rightarrow -3 = -3k \Rightarrow k = 1$. [1]
Polynomial is $x^2 - 2x - 3$. [1]

Q9. $y = x^2 - 3x - 4$.
Points: $(-1, 0), (4, 0), (0, -4)$. Vertex: $(1.5, -6.25)$.
Graph intersects x-axis at -1 and 4. Zeroes are -1 and 4. [3]

Q10. $p(x) = -x^2 + 2x + 3$.

(i) $a = -1 < 0$, so parabola opens downwards. [1]

(ii) $-x^2 + 2x + 3 = 0 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0$.
Zeroes: 3, -1. [1]

(iii) Vertex x-coord $= -b/2a = -2/-2 = 1$.
y-coord $= -(1)^2 + 2(1) + 3 = 4$. Vertex: (1, 4). [1]

(iv) Graph:

[2]

(i) The equation is quadratic with $a < 0$, so the shape is a Parabola opening downwards. [1]

(ii) Ball hits ground when $h(t) = 0$.
$-t^2 + 6t = 0 \Rightarrow t(6 - t) = 0$.
$t = 0$ (start) or $t = 6$. It hits ground at 6 seconds. [1]

(iii) Max height is at vertex. $t = -b/2a = -6/-2 = 3$ sec.
Max Height $h(3) = -(3)^2 + 6(3) = -9 + 18 = 9$ meters. [2]