Quadratic equations and functions

High School Math 1 · Unit 9 · quadratics-parabolas · Teacher edition

quadratics-parabolas

Name ______________________________Date ______________

1.
Solve $x^{2} - 5x + 6 = 0$ . Enter the larger solution.
Answer: ______________
2.
Solve $2x^{2} - 3x - 2 = 0$ using the quadratic formula. Enter the negative solution.
Answer: ______________
3.
A ball is thrown upward from a platform. Its height in feet after $t$ seconds is $h(t) = -16t^{2} + 64t + 80$ . What is the maximum height the ball reaches, in feet?
Answer: ______________
4.
What is the $x$ -coordinate of the vertex of $y = 3x^{2} - 12x + 7$ ?
1. (A)
  $-2$
2. (B)
  $2$
3. (C)
  $6$
4. (D)
  $12$
5.
The two real roots of $x^{2} - 7x + k = 0$ differ by $3$ . What is $k$ ?
1. (A)
  $4$
2. (B)
  $6$
3. (C)
  $8$
4. (D)
  $10$
5. (E)
  $12$

Answer key — Quadratic equations and functions

1.
3quadratics-parabolas-01
$(x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $x = 3$ . The larger solution is $\mathbf{3}$ .
2.
-1/2quadratics-parabolas-02
$x = \dfrac{3 \pm \sqrt{25}}{4} = \dfrac{3 \pm 5}{4}$ , so $x = 2$ or $x = \dfrac{-2}{4} = \mathbf{-\dfrac{1}{2}}$ . Factoring check: $(2x + 1)(x - 2) = 2x^2 - 3x - 2$ . ✓
3.
144quadratics-parabolas-03
Vertex at $t = -\frac{64}{-32} = 2$ . Then $h(2) = -16(4) + 64(2) + 80 = -64 + 128 + 80 = \mathbf{144}$ feet. ( $2$ is the time, not the height; $80$ is the starting height.)
4.
(B)
$2$
quadratics-parabolas-04
$x = -\dfrac{b}{2a} = -\dfrac{-12}{6} = \mathbf{2}$ . ( $6$ drops the $a$ and computes $-\frac{b}{2}$ ; $-2$ loses the sign; $12$ is just $-b$ .)
5.
(D)
$10$
quadratics-parabolas-05
$r + s = 7$ and $r - s = 3$ give $r = 5$ , $s = 2$ . So $k = rs = \mathbf{10}$ . Check: $x^2 - 7x + 10 = (x - 5)(x - 2)$ , roots $5$ and $2$ , which differ by $3$ . ✓ ( $12$ comes from guessing roots $3$ and $4$ , which differ by $1$ .)