Quadratic Functions

1. a) Complete the square:

$f(x) = x^2 - 6x + 8 = (x^2 - 6x + 9) - 9 + 8 = (x - 3)^2 - 1$

b) The vertex form is $f(x) = (x - 3)^2 - 1$, so:

• Vertex: $(3, -1)$
• Axis of symmetry: $x = 3$

c) Set $f(x) = 0$:

$x^2 - 6x + 8 = 0$ $(x - 2)(x - 4) = 0$

The $x$-intercepts are $x = 2$ and $x = 4$.

d) The graph is an upward-opening parabola with vertex $(3, -1)$, axis of symmetry $x = 3$, $x$-intercepts at $(2, 0)$ and $(4, 0)$, and $y$-intercept at $(0, 8)$.

2. a) With zeros at $x = -1$ and $x = 5$:

$f(x) = a(x + 1)(x - 5)$

b) Substitute the point $(1, -8)$:

$-8 = a(1 + 1)(1 - 5) = a(2)(-4) = -8a$

$a = 1$

c) Expand:

$f(x) = (x + 1)(x - 5) = x^2 - 5x + x - 5 = x^2 - 4x - 5$

Verification: $f(1) = 1 - 4 - 5 = -8$ $\checkmark$

3. a) The initial height is $h(0)$:

$h(0) = -16(0)^2 + 48(0) + 64 = 64 \text{ feet}$

b) The time of maximum height is at the vertex. For $h(t) = -16t^2 + 48t + 64$:

$t = -\frac{b}{2a} = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \text{ seconds}$

Maximum height:

$h(1.5) = -16(1.5)^2 + 48(1.5) + 64 = -16(2.25) + 72 + 64 = -36 + 72 + 64 = 100 \text{ feet}$

c) Set $h(t) = 0$:

$-16t^2 + 48t + 64 = 0$

Divide by $-16$:

$t^2 - 3t - 4 = 0$ $(t - 4)(t + 1) = 0$

$t = 4$ or $t = -1$. Since time must be positive, the ball hits the ground at $t = 4$ seconds.

Verification: $h(4) = -16(16) + 48(4) + 64 = -256 + 192 + 64 = 0$ $\checkmark$

4. The discriminant is $\Delta = b^2 - 4ac$.

a) $g(x) = 2x^2 + 3x + 5$:

$\Delta = 3^2 - 4(2)(5) = 9 - 40 = -31$

Since $\Delta < 0$, the function has no $x$-intercepts (the parabola does not cross the $x$-axis).

b) $h(x) = x^2 - 10x + 25$:

$\Delta = (-10)^2 - 4(1)(25) = 100 - 100 = 0$

Since $\Delta = 0$, the function has exactly one $x$-intercept (the vertex touches the $x$-axis). That intercept is $x = \frac{10}{2} = 5$.

c) $k(x) = -3x^2 + 7x - 1$:

$\Delta = 7^2 - 4(-3)(-1) = 49 - 12 = 37$

Since $\Delta > 0$, the function has two $x$-intercepts.

5. a) The vertex form is $q(x) = a(x - h)^2 + k$ with vertex $(h, k) = (2, -3)$:

$q(x) = a(x - 2)^2 - 3$

Substitute $(0, 5)$:

$5 = a(0 - 2)^2 - 3 = 4a - 3$ $4a = 8$ $a = 2$

So $q(x) = 2(x - 2)^2 - 3$.

b) Expand:

$q(x) = 2(x^2 - 4x + 4) - 3 = 2x^2 - 8x + 8 - 3 = 2x^2 - 8x + 5$

c) $q(6) = 2(6)^2 - 8(6) + 5 = 72 - 48 + 5 = 29$

Alternatively, using vertex form: $q(6) = 2(6 - 2)^2 - 3 = 2(16) - 3 = 32 - 3 = 29$ $\checkmark$