Problem 19

Number 19

Given the following graph:

Function Graph

Graph of linear and exponential functions.

point $A(0, -2)$
point $B(0, 6)$
point $C(-3, 0)$
point $D(-2, 2)$

The correct statements are

$1$ , $2$ , and $3$

$1$ and $3$

$2$ and $4$

$4$ only

$1$ , $2$ , $3$ , and $4$

Explanation

We will check the validity of each statement based on the given graph. The graph consists of the line $y = 2x + 6$ and the exponential curve $y = (1/2)^x - 3$ .

Analysis of Statement 1

Point A is the y-intercept of the curve $y = (1/2)^x - 3$ . Substitute $x = 0$ :

y = \left(\frac{1}{2}\right)^0 - 3 = 1 - 3 = -2

So, point A is $(0, -2)$ . Statement $1$ is correct.

Analysis of Statement 2

Point B is the y-intercept of the line $y = 2x + 6$ . Substitute $x = 0$ :

y = 2(0) + 6 = 6

So, point B is $(0, 6)$ . Statement $2$ is correct.

Analysis of Statement 3

Point C is the x-intercept of the line $y = 2x + 6$ . Substitute $y = 0$ :

\begin{aligned} 2x + 6 &= 0 \\ 2x &= -6 \\ x &= -3 \end{aligned}

So, point C is $(-3, 0)$ . Statement $3$ is correct.

Analysis of Statement 4

Statement $4$ states that point D is $(-2, 2)$ . On the graph, D is the intersection point of the line and the curve. We check if the point $(-2, 2)$ lies on both functions.

For the line $y = 2x + 6$ :

y = 2(-2) + 6 = -4 + 6 = 2

Point $(-2, 2)$ is on the line.

For the curve $y = (1/2)^x - 3$ :

y = \left(\frac{1}{2}\right)^{-2} - 3 = 2^2 - 3 = 4 - 3 = 1

Since $y = 1 \neq 2$ , point $(-2, 2)$ is not on the curve. Therefore, D is not $(-2, 2)$ . Statement $4$ is incorrect.

Thus, the correct statements are $1$ , $2$ , and $3$ .

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