Question

a quadratic function is shown.
$h(r) = -2(r + 3)(r - 1)$
complete the statements.
the function is
on the interval $r < -3$.
the function is positive and decreasing on the interval

Explanation:

Step1: Find the roots and vertex

The quadratic function is given in factored form \( h(r) = -2(r + 3)(r - 1) \). The roots are \( r=-3 \) and \( r = 1 \). The coefficient of \( r^2 \) is \( -2 \), so the parabola opens downward. The vertex (maximum point) is at the midpoint of the roots: \( r=\frac{-3 + 1}{2}=-1 \).

Step2: Analyze the interval \( r < -3 \)

For \( r < -3 \), let's pick a test point, say \( r=-4 \). Substitute into \( h(r) \): \( h(-4)=-2(-4 + 3)(-4 - 1)=-2(-1)(-5)=-10 \), which is negative. The function is decreasing before the vertex? Wait, no: since the parabola opens downward, the left side of the vertex (left of \( r=-1 \)): from \( r=-\infty \) to \( r=-1 \), the function is increasing (because it's going from \( -\infty \) up to the vertex). Wait, no: the derivative (or the slope) - the quadratic with negative leading coefficient: the axis of symmetry is \( r=-1 \). So for \( r < -1 \), the function is increasing (since moving from left to right towards the vertex, the function rises), and for \( r > -1 \), it's decreasing. But in \( r < -3 \), which is left of \( r=-3 \) (left of the left root), let's check the sign and the direction. Wait, when \( r < -3 \), both \( (r + 3) \) and \( (r - 1) \) are negative (since \( r < -3 \) implies \( r + 3 < 0 \) and \( r - 1 < 0 \)), so their product is positive, multiplied by \( -2 \) (negative) gives negative. Now, the slope: the derivative \( h'(r)=-2[(r - 1)+(r + 3)]=-2(2r + 2)=-4r - 4 \). For \( r < -3 \), \( h'(r)=-4r - 4 \). If \( r=-4 \), \( h'(-4)=16 - 4 = 12>0 \), so the function is increasing? Wait, no, earlier calculation with \( r=-4 \) gave \( h(-4)=-10 \), \( r=-5 \): \( h(-5)=-2(-5 + 3)(-5 - 1)=-2(-2)(-6)=-24 \), which is more negative. Wait, that contradicts the derivative. Wait, no, I made a mistake in the derivative. Let's recalculate the derivative: \( h(r)=-2(r + 3)(r - 1)=-2(r^2 - r + 3r - 3)=-2(r^2 + 2r - 3)=-2r^2 - 4r + 6 \). Then \( h'(r)=-4r - 4 \). For \( r < -3 \), say \( r=-4 \), \( h'(-4)=16 - 4 = 12>0 \), so the function is increasing. But the value at \( r=-4 \) is -10, \( r=-5 \) is -24, which is decreasing? Wait, no, that's a mistake. Wait, \( h(-4)=-2(-4 + 3)(-4 - 1)=-2(-1)(-5)=-10 \), \( h(-5)=-2(-5 + 3)(-5 - 1)=-2(-2)(-6)=-24 \). So as \( r \) decreases (goes from -4 to -5), \( h(r) \) decreases (from -10 to -24). But the derivative at \( r=-4 \) is positive, which means when \( r \) increases (from -5 to -4), \( h(r) \) increases (from -24 to -10). So in the interval \( r < -3 \), as \( r \) increases (moves to the right), the function increases (from more negative to less negative), but the function values are negative. So the function is negative and increasing? Wait, no, the options: the first blank options are about sign and direction. Wait, maybe I messed up. Wait, the roots are at -3 and 1, parabola opens downward. So the regions:

- \( r < -3 \): both factors \( (r + 3) \) and \( (r - 1) \) are negative, product positive, times -2: negative. The function is increasing (since moving from left to right towards the vertex at \( r=-1 \), the function rises from \( -\infty \) up to the vertex). Wait, but when \( r < -3 \), we are left of the left root. Let's take two points: \( r=-4 \) (h=-10) and \( r=-3.5 \) (h=-2(-0.5)(-4.5)=-2(2.25)=-4.5). So as \( r \) increases from -4 to -3.5, \( h(r) \) increases from -10 to -4.5 (less negative), so it's increasing. And the sign is negative. So the first blank: negative and increasing? Wait, no, the options include "negative and increasing". Wait, but let's check the next part.

Step3: Analyze the interval for positive and decreasing

The function is positive when the product \( (r + 3)(r - 1) \) is negative (since multiplied by -2, negative times negative is positive). So \( (r + 3)(r - 1) < 0 \) when \( -3 < r < 1 \) (because between the roots, the product of a positive and a negative is negative). Now, within \( -3 < r < 1 \), the function is decreasing when \( r > -1 \) (since the vertex is at \( r=-1 \), so to the right of \( r=-1 \) (between \( -1 \) and \( 1 \)), the function is decreasing (moving from the vertex down to \( r=1 \)). Wait, but we need positive and decreasing. Positive is in \( -3 < r < 1 \). Decreasing is in \( r > -1 \). So the intersection is \( -1 < r < 1 \)? Wait, no, let's check the derivative. For \( -3 < r < 1 \), the function is positive. The derivative \( h'(r)=-4r - 4 \). Set to zero: \( -4r - 4 = 0 \implies r=-1 \). So for \( r > -1 \), \( h'(r) < 0 \) (decreasing), for \( r < -1 \), \( h'(r) > 0 \) (increasing). So in \( -3 < r < 1 \), the function is increasing on \( -3 < r < -1 \) and decreasing on \( -1 < r < 1 \). And it's positive in \( -3 < r < 1 \). So positive and decreasing is on \( -1 < r < 1 \)? Wait, no, the options have \( -3 < r < -2 \), \( -1 < r < 1 \), etc. Wait, let's re-express:

Wait, the first statement: "The function is [blank] on the interval \( r < -3 \)". Let's recheck the sign and direction. For \( r < -3 \), as we saw, the function is negative (since \( h(r) = -2(r + 3)(r - 1) \), when \( r < -3 \), \( (r + 3) < 0 \), \( (r - 1) < 0 \), so product positive, times -2: negative). The direction: since the vertex is at \( r=-1 \), moving from \( r=-\infty \) to \( r=-1 \), the function is increasing (because the parabola opens downward, so left side of vertex is increasing, right side is decreasing). So for \( r < -3 \) (left of \( r=-3 \), which is left of the vertex's left side), the function is increasing (since moving towards the vertex, which is at \( r=-1 \), so from \( r=-\infty \) to \( r=-1 \), it's increasing). So in \( r < -3 \), the function is negative and increasing? Wait, but the options for the first blank: "negative and increasing", "negative and decreasing", etc. Wait, maybe I made a mistake earlier. Let's take \( r=-5 \) and \( r=-4 \): \( h(-5)=-2(-5+3)(-5-1)=-2(-2)(-6)=-24 \), \( h(-4)=-2(-4+3)(-4-1)=-2(-1)(-5)=-10 \). So as \( r \) increases from -5 to -4 (moving right), \( h(r) \) increases from -24 to -10 (less negative), so it's increasing. And it's negative. So the first blank is "negative and increasing".

Now the second statement: "The function is positive and decreasing on the interval [blank]". Positive is in \( -3 < r < 1 \). Decreasing is when \( r > -1 \) (since derivative is negative when \( r > -1 \): \( h'(r)=-4r - 4 < 0 \implies -4r < 4 \implies r > -1 \)). So the interval where \( -3 < r < 1 \) and \( r > -1 \) is \( -1 < r < 1 \). Wait, but let's check the options. The options for the interval include \( -1 < r < 1 \).

Wait, but let's re-examine the first part. Wait, maybe I messed up the direction. Let's think again: the parabola opens downward, vertex at \( r=-1 \). So the left arm (from \( -\infty \) to \( r=-1 \)) is increasing (going up to the vertex), the right arm (from \( r=-1 \) to \( \infty \)) is decreasing (going down from the vertex). So for \( r < -3 \) (left of \( r=-3 \), which is left of the vertex's left side), the function is increasing (since moving towards the vertex, which is to the right). And it's negative (as we saw). So first blank: negative and increasing.

For the second part: positive is between \( -3 \) and \( 1 \). Decreasing is on the right arm, which is \( r > -1 \). So the overlap is \( -1 < r < 1 \), where the function is positive (between roots) and decreasing (right of vertex).

Wait, but let's check the options. The first blank options: "negative and increasing" is an option. The second blank options: \( -1 < r < 1 \) is an option.

Answer:

The function is \(\boldsymbol{\text{negative and increasing}}\) on the interval \( r < -3 \).
The function is positive and decreasing on the interval \(\boldsymbol{-1 < r < 1}\).