The problem asks us to sketch the graphs of the following quadratic functions: (a) $f(x) = x^2 - 2x - 3$ (b) $f(x) = x^2 - 9$ (c) $f(x) = -2x^2 + 8x$ (d) $f(x) = -x^2 - 3x + 4$

1. Problem Description

The problem asks us to sketch the graphs of the following quadratic functions:

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

(b) $f(x) = x^2 - 9$

(c) $f(x) = -2x^2 + 8x$

(d) $f(x) = -x^2 - 3x + 4$

2. Solution Steps

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

To sketch the graph, we can find the vertex, the axis of symmetry, and the x-intercepts.

The axis of symmetry is given by $x = -\frac{b}{2a}$. Here, $a=1$ and $b=-2$, so $x = -\frac{-2}{2(1)} = 1$.

The vertex has an x-coordinate of 1, so its y-coordinate is $f(1) = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4$. Thus, the vertex is $(1, -4)$.

The x-intercepts are the solutions to $x^2 - 2x - 3 = 0$, which factors as $(x-3)(x+1) = 0$. Thus, the x-intercepts are $x = 3$ and $x = -1$.

(b) $f(x) = x^2 - 9$

The axis of symmetry is $x = -\frac{b}{2a}$. Here, $a=1$ and $b=0$, so $x = -\frac{0}{2(1)} = 0$.

The vertex has an x-coordinate of 0, so its y-coordinate is $f(0) = (0)^2 - 9 = -9$. Thus, the vertex is $(0, -9)$.

The x-intercepts are the solutions to $x^2 - 9 = 0$, so $x^2 = 9$, which means $x = \pm 3$. Thus, the x-intercepts are $x = 3$ and $x = -3$.

(c) $f(x) = -2x^2 + 8x$

The axis of symmetry is $x = -\frac{b}{2a}$. Here, $a=-2$ and $b=8$, so $x = -\frac{8}{2(-2)} = 2$.

The vertex has an x-coordinate of 2, so its y-coordinate is $f(2) = -2(2)^2 + 8(2) = -2(4) + 16 = -8 + 16 = 8$. Thus, the vertex is $(2, 8)$.

The x-intercepts are the solutions to $-2x^2 + 8x = 0$, which factors as $-2x(x-4) = 0$. Thus, the x-intercepts are $x = 0$ and $x = 4$.

(d) $f(x) = -x^2 - 3x + 4$

The axis of symmetry is $x = -\frac{b}{2a}$. Here, $a=-1$ and $b=-3$, so $x = -\frac{-3}{2(-1)} = -\frac{3}{2} = -1.5$.

The vertex has an x-coordinate of -1.5, so its y-coordinate is $f(-1.5) = -(-1.5)^2 - 3(-1.5) + 4 = -2.25 + 4.5 + 4 = 6.25$. Thus, the vertex is $(-1.5, 6.25)$.

The x-intercepts are the solutions to $-x^2 - 3x + 4 = 0$, which can be multiplied by -1 to get $x^2 + 3x - 4 = 0$. This factors as $(x+4)(x-1) = 0$. Thus, the x-intercepts are $x = -4$ and $x = 1$.

3. Final Answer

The solutions involve sketching the following parabolas:

(a) $f(x) = x^2 - 2x - 3$, vertex at $(1, -4)$, x-intercepts at $x=3$ and $x=-1$.

(b) $f(x) = x^2 - 9$, vertex at $(0, -9)$, x-intercepts at $x=3$ and $x=-3$.

(c) $f(x) = -2x^2 + 8x$, vertex at $(2, 8)$, x-intercepts at $x=0$ and $x=4$.

(d) $f(x) = -x^2 - 3x + 4$, vertex at $(-1.5, 6.25)$, x-intercepts at $x=-4$ and $x=1$.

Note: A sketch is expected for each function, plotting these features and drawing a smooth curve.