Question

consider the graph of the quadratic function. which interval on the x-axis has a negative rate of change? -2 to -1 -1.5 to 0 0 to 1 1 to 2.5

Explanation:

Step1: Understand Rate of Change

The rate of change of a function on an interval is positive if the function increases (y - values rise as x increases) and negative if it decreases (y - values fall as x increases). For a quadratic function (parabola) opening downward (as this one does, since the vertex is the maximum point), the function increases before the vertex and decreases after the vertex. The vertex here is at \(x = 0\) (from the graph, the peak is at \(x = 0\)).

Step2: Analyze Each Interval

- **Interval - 2 to - 1**: As \(x\) increases from - 2 to - 1, the \(y\) - values are increasing (moving from left to right towards the vertex), so the rate of change is positive.
- **Interval - 1.5 to 0**: As \(x\) increases from - 1.5 to 0, the \(y\) - values are increasing (moving towards the vertex), so the rate of change is positive.
- **Interval 0 to 1**: As \(x\) increases from 0 to 1, the \(y\) - values start to decrease (moving away from the vertex to the right), but let's check the next interval too.
- **Interval 1 to 2.5**: As \(x\) increases from 1 to 2.5, the \(y\) - values are decreasing (since the parabola is opening downward and we are to the right of the vertex \(x = 0\), the function is decreasing here). Also, between 0 to 1, the function is still decreasing but let's confirm the slope. The rate of change is \(\frac{\Delta y}{\Delta x}\). For the interval 1 to 2.5, as \(x\) increases, \(y\) decreases, so \(\Delta y\) is negative and \(\Delta x\) is positive, so the rate of change is negative. For 0 to 1, let's take two points: at \(x = 0\), \(y = 3\); at \(x = 1\), let's estimate \(y\) - value. From the graph, at \(x = 1\), \(y\) is between 2 and 3? Wait, no, the vertex is at (0, 3). At \(x = 1\), the \(y\) - value: looking at the graph, the parabola passes through ( - 1, 0) and (2, 0). The equation of the parabola: since roots are \(x=-1\) and \(x = 2\), the equation is \(y=a(x + 1)(x - 2)\). At \(x = 0\), \(y=3\), so \(3=a(1)(-2)\), so \(a=-\frac{3}{2}\). So \(y =-\frac{3}{2}(x + 1)(x - 2)\). At \(x = 1\), \(y=-\frac{3}{2}(2)(-1)=3\)? Wait, no, that can't be. Wait, maybe my estimation is wrong. Wait, the vertex is at (0, 3). So the parabola is symmetric about \(x=\frac{-1 + 2}{2}=0.5\)? Wait, no, the vertex form of a parabola is \(y=a(x - h)^2+k\), where \((h,k)\) is the vertex. Here, vertex is (0, 3), so \(y=a(x - 0)^2+3=ax^{2}+3\). It passes through ( - 1, 0), so \(0=a(-1)^{2}+3\), so \(a=-3\). So the equation is \(y=-3x^{2}+3\). Now, at \(x = 0\), \(y = 3\); at \(x = 1\), \(y=-3(1)+3 = 0\)? Wait, no, that's not matching the graph. Wait, the graph shows that at \(x = 1\), the \(y\) - value is above 0. Oh, I must have misread the roots. The graph crosses the x - axis at \(x=-1\) and \(x = 2\)? Wait, no, looking at the graph, the left root is at \(x=-1\) (since at \(x=-1\), \(y = 0\)) and the right root is at \(x = 2\) (at \(x = 2\), \(y = 0\)). The vertex is at (0, 3). So the equation is \(y=-3x^{2}+3\) (since when \(x = 0\), \(y = 3\); when \(x = 1\), \(y=-3 + 3=0\)? But in the graph, at \(x = 1\), the \(y\) - value is not zero. Wait, maybe the roots are at \(x=-1\) and \(x = 1\)? No, the graph shows that at \(x = 2\), \(y = 0\). Wait, maybe my initial analysis of the roots is wrong. Anyway, the key point is that the parabola is opening downward with vertex at (0, 3). So to the right of the vertex (x > 0), the function is decreasing. So the interval where \(x\) is increasing and \(y\) is decreasing (negative rate of change) is when \(x\) is to the right of the vertex. The vertex is at \(x = 0\), so intervals where \(x\) increases and \(y\) decreases are \(x>0\). Among the options, 0 to 1 and 1 to 2.5. But let's calculate the rate of change for 0 to 1: at \(x = 0\), \(y = 3\); at \(x = 1\), using the equation \(y=-3x^{2}+3\), \(y = 0\)? But in the graph, at \(x = 1\), the \(y\) - value is above 0. Wait, maybe the vertex is at (0, 3), and the parabola passes through ( - 1, 0) and (2, 0). So the axis of symmetry is \(x=\frac{-1 + 2}{2}=0.5\). Ah! I made a mistake earlier. The axis of symmetry is at \(x = 0.5\), not \(x = 0\). So the vertex is at \(x = 0.5\), \(y\) - value: let's calculate. The equation is \(y=a(x + 1)(x - 2)\). At \(x = 0.5\), \(y=a(1.5)(-1.5)=-2.25a\). The vertex is the maximum point, so we can find \(a\) by using the fact that at \(x = 0\), \(y = 3\) (from the graph, the point (0, 3) is on the parabola). So \(3=a(1)(-2)\), so \(a=-\frac{3}{2}\). Then at \(x = 0.5\), \(y=-\frac{3}{2}(1.5)(-1.5)=-\frac{3}{2}\times(-2.25)=3.375\). So the vertex is at (0.5, 3.375). Now, let's re - analyze the intervals:
- **Interval - 2 to - 1**: As \(x\) increases from - 2 to - 1, moving towards the axis of symmetry (\(x = 0.5\)), so \(y\) - values are increasing (rate of change positive).
- **Interval - 1.5 to 0**: As \(x\) increases from - 1.5 to 0, moving towards the axis of symmetry, \(y\) - values are increasing (rate of change positive).
- **Interval 0 to 1**: As \(x\) increases from 0 to 1, we cross the axis of symmetry (\(x = 0.5\)). From \(x = 0\) to \(x = 0.5\), \(y\) - values increase (towards vertex), and from \(x = 0.5\) to \(x = 1\), \(y\) - values decrease (away from vertex). So the rate of change in 0 to 1: let's take \(x = 0\), \(y = 3\); \(x = 1\), \(y\): using the equation \(y=-\frac{3}{2}(x + 1)(x - 2)\), at \(x = 1\), \(y=-\frac{3}{2}(2)(-1)=3\). Wait, so at \(x = 0\), \(y = 3\); at \(x = 1\), \(y = 3\)? No, that's not right. Wait, when \(x = 0\), \(y=-\frac{3}{2}(1)(-2)=3\); when \(x = 1\), \(y=-\frac{3}{2}(2)(-1)=3\); when \(x = 0.5\), \(y=-\frac{3}{2}(1.5)(-1.5)=3.375\). So from \(x = 0\) to \(x = 0.5\), \(y\) increases from 3 to 3.375 (rate of change positive: \(\frac{3.375 - 3}{0.5 - 0}=\frac{0.375}{0.5}=0.75\)). From \(x = 0.5\) to \(x = 1\), \(y\) decreases from 3.375 to 3 (rate of change negative: \(\frac{3 - 3.375}{1 - 0.5}=\frac{-0.375}{0.5}=-0.75\)). So the interval 0 to 1 has a mixed rate of change (positive then negative), but the overall average? Wait, the question is about the interval having a negative rate of change. Let's check the interval 1 to 2.5:
At \(x = 1\), \(y = 3\); at \(x = 2.5\), let's calculate \(y\): \(y=-\frac{3}{2}(2.5 + 1)(2.5 - 2)=-\frac{3}{2}(3.5)(0.5)=-\frac{3}{2}\times1.75=-2.625\). So from \(x = 1\) to \(x = 2.5\), \(x\) increases by \(1.5\), \(y\) decreases from 3 to - 2.625. So \(\Delta y=-2.625 - 3=-5.625\), \(\Delta x=2.5 - 1 = 1.5\), rate of change is \(\frac{-5.625}{1.5}=-3.75\) (negative). For the interval 0 to 1: \(\Delta x = 1\), \(\Delta y=3 - 3 = 0\) (wait, no, at \(x = 0\), \(y = 3\); at \(x = 1\), \(y = 3\), so \(\Delta y = 0\), rate of change 0. But that's not possible. I think my initial graph reading was wrong. The correct way is: the parabola is a downward - opening parabola with vertex at (0, 3) (as per the graph, the peak is at (0, 3)). So the axis of symmetry is \(x = 0\). Then the roots are symmetric about \(x = 0\). But the graph shows it crossing the x - axis at \(x=-1\) and \(x = 1\)? Wait, that makes more sense. If the vertex is at (0, 3) and it crosses the x - axis at \(x=-1\) and \(x = 1\), then the equation is \(y=-3x^{2}+3\) (since at \(x = 1\), \(y=-3 + 3 = 0\); at \(x = 0\), \(y = 3\)). Then, the axis of symmetry is \(x = 0\). Now, let's re - analyze:
- **Interval - 2 to - 1**: As \(x\) increases from - 2 to - 1, moving towards \(x = 0\) (axis of symmetry), \(y\) - values increase (from \(y=-3(-2)^{2}+3=-9\) to \(y=-3(-1)^{2}+3 = 0\)), so rate of change is positive (\(\frac{0-(-9)}{-1-(-2)}=\frac{9}{1}=9>0\)).
- **Interval - 1.5 to 0**: As \(x\) increases from - 1.5 to 0, \(y\) - values increase (from \(y=-3(-1.5)^{2}+3=-3\times2.25 + 3=-6.75 + 3=-3.75\) to \(y = 3\)), rate of change is positive (\(\frac{3-(-3.75)}{0-(-1.5)}=\frac{6.75}{1.5}=4.5>0\)).
- **Interval 0 to 1**: As \(x\) increases from 0 to 1, \(y\) - values decrease (from \(y = 3\) to \(y=-3(1)^{2}+3 = 0\)), rate of change is \(\frac{0 - 3}{1 - 0}=-3<0\). Wait, but earlier I thought it was positive. But according to the equation \(y=-3x^{2}+3\), at \(x = 0\), \(y = 3\); at \(x = 1\), \(y = 0\). So the rate of change is - 3.
- **Interval 1 to 2.5**: As \(x\) increases from 1 to 2.5, \(y\) - values decrease (from \(y = 0\) to \(y=-3(2.5)^{2}+3=-3\times6.25+3=-18.75 + 3=-15.75\)), rate of change is \(\frac{-15.75-0}{2.5 - 1}=\frac{-15.75}{1.5}=-10.5<0\).

Now, the question is which interval has a negative rate of change. Both 0 to 1 and 1 to 2.5 have negative rate of change. But let's check the options again. Wait, maybe the graph is as I first misread, with vertex at (0, 3) and roots at ( - 1, 0) and (2, 0). Then the axis of symmetry is \(x=\frac{-1 + 2}{2}=0.5\). Let's take points:
- At \(x=-2\), \(y\): using \(y=a(x + 1)(x - 2)\), at \(x = 0\), \(y = 3\), so \(3=a(1)(-2)\), \(a=-\frac{3}{2}\). At \(x=-2\), \(y=-\frac{3}{2}(-1)(-4)=-\frac{3}{2}\times4=-6\).
- At \(x=-1\), \(y = 0\).
- At \(x = 0\), \(y = 3\).
- At \(x = 1\), \(y=-\frac{3}{2}(2)(-1)=3\).
- At \(x = 2\), \(y = 0\).
- At \(x = 2.5\), \(y=-\frac{3}{2}(3.5)(0.5)=-\frac{3}{2}\times1.75=-2.625\).

Now, rate of change for each interval:
- **-2 to -1**: \(\frac{0-(-6)}{-1-(-2)}=\frac{6}{1}=6>0\) (positive).
- **-1.5 to 0**: At \(x=-1.5\), \(y=-\frac{3}{2}(-0.5)(-3.5)=-\frac{3}{2}\times1.75=-2.625\); at \(x = 0\), \(y = 3\). Rate of change: \(\frac{3-(-2.625)}{0-(-1.5)}=\frac{5.625}{1.5}=3.75>0\) (positive).
- **0 to 1**: At \(x = 0\), \(y = 3\); at \(x = 1\), \(y = 3\). Rate of change: \(\frac{3 - 3}{1 - 0}=0\) (zero).
- **1 to 2.5**: At \(x = 1\), \(y = 3\); at \(x = 2.5\), \(y=-2.625\). Rate of change: \(\frac{-2.625 - 3}{2.5 - 1}=\frac{-5.625}{1.5}=-3.75<0\) (negative).

So the interval with negative rate of change is 1 to 2.5.

Answer:

D. 1 to 2.5 (assuming the options are labeled as A: - 2 to - 1, B: - 1.5 to 0, C: 0 to 1, D: 1 to 2.5)