Question

the function $y = f(x)$ is graphed below. what is the average rate of change of the function $f(x)$ on the interval $0 \leq x \leq 3$?

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \( f(x) \) on the interval \([a, b]\) is given by \(\frac{f(b) - f(a)}{b - a}\). Here, \( a = 0 \) and \( b = 3 \), so we need to find \( f(0) \) and \( f(3) \) from the graph.

Step2: Find \( f(0) \) from the graph

Looking at the graph, when \( x = 0 \), the \( y \)-coordinate (which is \( f(0) \)) is \(-15\) (we can see the point on the \( y \)-axis around \( y=-15\)).

Step3: Find \( f(3) \) from the graph

When \( x = 3 \), we look at the graph. From the graph, at \( x = 3 \), the \( y \)-coordinate ( \( f(3) \)) is \(-20\)? Wait, no, wait. Wait, the right - hand parabola: let's check the points. Wait, maybe I made a mistake. Wait, the graph: when \( x = 0 \), the point is at \( y=-15\)? Wait, no, looking at the grid, each square is, let's see, the \( y \)-axis: from \( -50 \) to \( 50 \), with each grid line maybe 5 units? Wait, no, let's re - examine. Wait, the left part: when \( x = 0 \), the point is at \( y=-15\)? Wait, no, the graph at \( x = 0 \): the point is at \( (0, - 15) \)? Wait, no, maybe the grid is 5 units per square? Wait, no, let's check the right parabola. At \( x = 3 \), what's the \( y \)-value? Wait, the right parabola has a root at \( x = 3 \)? No, the root is at \( x = 3 \)? Wait, the right parabola crosses the \( x \)-axis at \( x = 3 \)? Wait, no, the graph: the right parabola, when \( x = 3 \), is it on the \( x \)-axis? No, the root is at \( x = 3 \)? Wait, no, the point at \( x = 3 \): looking at the graph, the right parabola has a minimum? Wait, no, the graph: let's list the points. At \( x = 0 \), the \( y \)-coordinate is \(-15\) (let's assume each grid square is 5 units: from \( y=-20\) to \( y=-10\), so \( x = 0 \) is at \( y=-15\)). At \( x = 3 \), let's see, the right parabola: when \( x = 3 \), what's \( f(3) \)? Wait, maybe I misread. Wait, the formula is \(\frac{f(3)-f(0)}{3 - 0}\). Let's find \( f(0) \) and \( f(3) \) correctly.

Wait, looking at the graph again: when \( x = 0 \), the point is at \( y=-15\) (so \( f(0)=-15\)). When \( x = 3 \), the point on the right parabola: let's see, the right parabola has a point at \( x = 3 \), what's its \( y \)-value? Wait, maybe the grid is 5 units. Wait, the bottom of the right parabola: the minimum point is at \( x = 1 \) or \( x = 2 \), with \( y=-25\)? No, maybe I made a mistake. Wait, let's check the average rate of change formula again. The average rate of change is \(\frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b - a}\).

Wait, let's look at the graph again. At \( x = 0 \), the \( y \)-coordinate is \(-15\) (so \( f(0)=-15\)). At \( x = 3 \), the \( y \)-coordinate: looking at the graph, when \( x = 3 \), the point is at \( y=-20\)? No, wait, maybe the correct values are: from the graph, at \( x = 0 \), \( f(0)=-15 \), and at \( x = 3 \), \( f(3)=-20 \)? No, that can't be. Wait, maybe I messed up. Wait, let's check the graph again. The left part: the curve crosses the \( y \)-axis at \( (0, - 15) \). The right part: at \( x = 3 \), what's the \( y \)-value? Wait, the right parabola, when \( x = 3 \), is it on the \( x \)-axis? No, the root is at \( x = 3 \)? Wait, the graph shows that the right parabola crosses the \( x \)-axis at \( x = 3 \)? No, the \( x \)-intercept is at \( x = 3 \), so \( f(3)=0 \)? Wait, that's a mistake earlier. Let's re - examine the graph. The right parabola: it crosses the \( x \)-axis at \( x = 3 \), so \( f(3)=0 \). And at \( x = 0 \), the \( y \)-coordinate is \(-15\) (so \( f(0)=-15\)).

Now, apply the average rate of change formula: \(\frac{f(3)-f(0)}{3 - 0}=\frac{0-(-15)}{3}=\frac{15}{3}=5\)? Wait, no, that doesn't seem right. Wait, maybe \( f(0) \) is \(-15\) and \( f(3) \) is \( 0 \)? Wait, let's check the graph again. The \( y \)-axis: the point at \( x = 0 \) is at \( y=-15 \) (since it's between \( y=-20 \) and \( y=-10 \), 15 units below the origin). The right parabola: at \( x = 3 \), it's on the \( x \)-axis, so \( y = 0 \). Then the average rate of change is \(\frac{0 - (-15)}{3-0}=\frac{15}{3}=5\)? Wait, but maybe I got \( f(0) \) wrong. Wait, maybe the point at \( x = 0 \) is \( f(0)=-15 \), and at \( x = 3 \), \( f(3)=0 \). Then the calculation is \(\frac{0 - (-15)}{3}=5\). Wait, but let's check again.

Wait, another way: the average rate of change is the slope of the secant line between \( (0,f(0)) \) and \( (3,f(3)) \). From the graph, \( (0, - 15) \) and \( (3,0) \). So the slope is \(\frac{0 - (-15)}{3-0}=\frac{15}{3}=5\). Wait, but maybe the \( y \)-value at \( x = 0 \) is \(-15\) and at \( x = 3 \) is \( 0 \). So the average rate of change is 5.

Wait, no, maybe I made a mistake in \( f(0) \). Let's look at the graph again. The curve at \( x = 0 \): the point is at \( y=-15 \) (each grid square is 5 units: from \( y=-20 \) (which is 4 squares down from \( y = 0 \)) to \( y=-10 \) (2 squares down), so \( x = 0 \) is at 3 squares down, so \( y=-15 \)). At \( x = 3 \), the point is at \( y = 0 \) (on the \( x \)-axis). So the average rate of change is \(\frac{f(3)-f(0)}{3 - 0}=\frac{0-(-15)}{3}=5\).

Wait, but let's confirm the formula. The average rate of change of a function \( y = f(x) \) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). So with \( a = 0 \), \( b = 3 \), we need \( f(0) \) and \( f(3) \). From the graph, \( f(0)=-15 \) and \( f(3)=0 \). Then \(\frac{0-(-15)}{3-0}=\frac{15}{3}=5\).

Answer:

5