Question

information about three functions is given below. use this information to complete each part.
function 1
the function is a quadratic function. the maximum output is -3, which occurs for an input of 0. the function passes through (1, -4).
function 2
the function is a linear function. the table gives some points on its graph.

x	-2	-1	0	1	2
y	-12	-9	-6	-3	0

function 3
the equation of the function is as follows.
$y=-(3)^x$
(a) choose the graph of each function.
function 1
(choose one)
function 2
(choose one)
function 3
(choose one)

Explanation:

Step1: Analyze Function 1

Function 1 is a quadratic function with maximum output -3 at \( x = 0 \), so its vertex is \( (0, -3) \) and it opens downward (since it has a maximum). It also passes through \( (1, -4) \). The general form of a quadratic function with vertex \( (h, k) \) is \( y = a(x - h)^2 + k \). Substituting \( h = 0 \), \( k = -3 \), and the point \( (1, -4) \):
\( -4 = a(1 - 0)^2 - 3 \)
\( -4 = a - 3 \)
\( a = -1 \). So the equation is \( y = -x^2 - 3 \). The graph is a parabola opening downward with vertex at \( (0, -3) \).

Step2: Analyze Function 2

Function 2 is linear. From the table, the slope \( m \) is \( \frac{-9 - (-12)}{-1 - (-2)} = \frac{3}{1} = 3 \), or using \( (0, -6) \) and \( (1, -3) \): \( m = \frac{-3 - (-6)}{1 - 0} = 3 \). The equation is \( y = 3x - 6 \) (since when \( x = 0 \), \( y = -6 \)). It's a line with slope 3, y-intercept -6, passing through the given points.

Step3: Analyze Function 3

Function 3 is \( y = -(3)^x \), an exponential function. When \( x = 0 \), \( y = -1 \); as \( x \) increases, \( 3^x \) grows, so \( y \) becomes more negative (but wait, the given equation is \( y = -3^x \)? Wait, the table for Function 2 has \( y \) increasing by 3, but Function 3: let's check \( x = 0 \): \( y = -3^0 = -1 \); \( x = 1 \): \( y = -3^1 = -3 \); \( x = 2 \): \( y = -3^2 = -9 \). Wait, but the table for Function 2 has \( y = 0 \) at \( x = 2 \), so Function 3's graph is exponential, decreasing (since the coefficient is negative, it's a reflection of \( 3^x \) over the x-axis, so it decreases as \( x \) increases? Wait no: \( 3^x \) increases, so \( -3^x \) decreases (goes to \( -\infty \)) as \( x \) increases, and approaches 0 as \( x \to -\infty \).

Now, for graph selection:

• Function 1: parabola opening downward, vertex (0, -3).
• Function 2: line with slope 3, y-intercept -6 (passes through (0, -6), (1, -3), etc.).
• Function 3: exponential, \( y = -3^x \), so at \( x = 0 \), \( y = -1 \); \( x = 1 \), \( y = -3 \); \( x = 2 \), \( y = -9 \), decreasing as \( x \) increases.

(Note: Since the graphs are not fully visible, but based on analysis:

• Function 1: parabola (quadratic) with vertex (0, -3), opening down.
• Function 2: line with positive slope, passing through the table points.
• Function 3: exponential curve, decreasing, passing through (0, -1), (1, -3), etc.)

Answer:

(Assuming the graphs are labeled, but since they are not shown, the process is to match each function's type to its graph:

• Function 1: Parabola (quadratic) with vertex (0, -3), opening downward.
• Function 2: Line with slope 3, y-intercept -6.
• Function 3: Exponential curve \( y = -3^x \).)

(If the graphs are, e.g., left: parabola, middle: line, right: exponential, then:
Function 1: Left graph (parabola), Function 2: Middle graph (line), Function 3: Right graph (exponential), but this depends on the actual graph shapes. The key is to match the function type to the graph: quadratic (parabola), linear (line), exponential (curve).)