Question

1. the graph of a quadratic function ( y = f(x) ) passes through ( (5, 1) ), ( (7, k) ), and ( (9, 26) ). the average rates of change over equal intervals of size 1 of ( f ) are changing at a constant rate of 1.5. find the value of ( k ).

Explanation:

Step1: Recall average rate of change formula

The average rate of change of a function \( y = f(x) \) over the interval \([a, b]\) is given by \( \frac{f(b)-f(a)}{b - a} \).

Step2: Calculate average rate from \( x = 5 \) to \( x = 7 \)

The interval from \( 5 \) to \( 7 \) has length \( 7 - 5=2 \), but the problem states the average rate over equal intervals of size \( 1 \) is constant at \( 1.5 \). Wait, actually, let's consider the intervals: from \( 5 \) to \( 7 \) is two intervals of size \( 1 \) (5 to 6, 6 to 7), and from \( 7 \) to \( 9 \) is also two intervals of size \( 1 \) (7 to 8, 8 to 9). But maybe a better approach: the average rate of change from \( x = 5 \) to \( x = 7 \) should be related to the average rate over each unit interval. Let's denote the average rate of change over \([5,6]\) as \( 1.5 \) and over \([6,7]\) as \( 1.5 \), and over \([7,8]\) as \( 1.5 \), over \([8,9]\) as \( 1.5 \).

First, find the average rate of change from \( x = 5 \) to \( x = 7 \): the number of intervals of size \( 1 \) is \( 7 - 5 = 2 \), so the total change from \( x = 5 \) to \( x = 7 \) is \( 1.5\times2 = 3 \). So \( f(7)-f(5)=3 \). We know \( f(5) = 1 \), so \( k - 1=3 \)? Wait, no, wait. Wait, also, from \( x = 7 \) to \( x = 9 \), the number of intervals of size \( 1 \) is \( 9 - 7 = 2 \), so the total change from \( x = 7 \) to \( x = 9 \) is \( 1.5\times2 = 3 \)? Wait, but \( f(9)=26 \), so \( f(9)-f(7)=3 \)? Then \( 26 - k = 3 \), so \( k = 23 \)? Wait, that can't be. Wait, maybe I messed up. Wait, the average rate of change over each interval of size \( 1 \) is \( 1.5 \). So the average rate of change from \( x = 5 \) to \( x = 6 \) is \( \frac{f(6)-f(5)}{6 - 5}=1.5 \), so \( f(6)=f(5)+1.5 = 1 + 1.5 = 2.5 \). Then from \( x = 6 \) to \( x = 7 \), average rate is \( 1.5 \), so \( f(7)=f(6)+1.5 = 2.5 + 1.5 = 4 \)? Wait, no, that contradicts the other way. Wait, maybe the quadratic function has a constant second difference? Wait, for a quadratic function, the second difference is constant. Let's denote the values at \( x = 5, 6, 7, 8, 9 \) as \( f(5)=1 \), \( f(6)=a \), \( f(7)=k \), \( f(8)=b \), \( f(9)=26 \). The first differences (average rate over interval of size 1) are \( a - 1 \), \( k - a \), \( b - k \), \( 26 - b \). The second differences are \( (k - a)-(a - 1)=k - 2a + 1 \), \( (b - k)-(k - a)=b - 2k + a \), \( (26 - b)-(b - k)=26 - 2b + k \). Since it's a quadratic function, the second differences are equal. Also, the average rate over interval of size 1 is constant? Wait, no, the average rate of change over equal intervals of size 1 is changing at a constant rate? Wait, the problem says "The average rates of change over equal intervals of size 1 of \( f \) are changing at a constant rate of 1.5". Wait, that means the first differences (average rate over interval of size 1) have a constant rate of change (i.e., the second differences are constant and equal to 1.5). Wait, let's clarify:

Let \( r_1 \) be the average rate from \( 5 \) to \( 6 \): \( r_1=\frac{f(6)-f(5)}{1}=f(6)-1 \)

\( r_2 \) be the average rate from \( 6 \) to \( 7 \): \( r_2=\frac{f(7)-f(6)}{1}=k - f(6) \)

\( r_3 \) be the average rate from \( 7 \) to \( 8 \): \( r_3=\frac{f(8)-f(7)}{1}=f(8)-k \)

\( r_4 \) be the average rate from \( 8 \) to \( 9 \): \( r_4=\frac{f(9)-f(8)}{1}=26 - f(8) \)

The problem states that the average rates of change (the \( r_i \)) are changing at a constant rate of \( 1.5 \). That means the difference between consecutive \( r_i \) is \( 1.5 \). So \( r_2 - r_1 = 1.5 \), \( r_3 - r_2 = 1.5 \), \( r_4 - r_3 = 1.5 \).

So let's denote \( r_1 = r \), then \( r_2 = r + 1.5 \), \( r_3 = r + 3 \), \( r_4 = r + 4.5 \)

Now, the total change from \( x = 5 \) to \( x = 9 \) is \( f(9)-f(5)=26 - 1 = 25 \)

But the total change is also the sum of the average rates times the interval size (which is 1 for each interval, so sum of \( r_1 + r_2 + r_3 + r_4 \))

So \( r + (r + 1.5) + (r + 3) + (r + 4.5) = 25 \)

Simplify: \( 4r + 9 = 25 \)

\( 4r = 16 \)

\( r = 4 \)

So \( r_1 = 4 \), \( r_2 = 5.5 \), \( r_3 = 7 \), \( r_4 = 8.5 \)

Now, \( r_1 = f(6)-1 = 4 \implies f(6)=5 \)

\( r_2 = k - f(6) = 5.5 \implies k - 5 = 5.5 \implies k = 10.5 \)? Wait, no, that can't be. Wait, maybe I misinterpreted the problem. Wait, the problem says "The average rates of change over equal intervals of size 1 of \( f \) are changing at a constant rate of 1.5". Maybe it's that the average rate of change over each interval of size 1 has a constant rate of change, i.e., the first derivative (if it were differentiable) has a constant rate of change, meaning the second derivative is constant (which is true for quadratic functions, since the second derivative of a quadratic \( ax^2 + bx + c \) is \( 2a \), constant). Wait, for a quadratic function \( f(x)=ax^2 + bx + c \), the average rate of change over \([x, x + 1]\) is \( f(x + 1)-f(x)=a(x + 1)^2 + b(x + 1) + c - (ax^2 + bx + c)=2ax + a + b \). So this is a linear function in \( x \), so the rate of change of the average rate of change (i.e., the derivative of \( 2ax + a + b \)) is \( 2a \), which is constant. So the problem says this rate of change is \( 1.5 \), so \( 2a = 1.5 \implies a = 0.75 \)

Now, we know \( f(5)=1 \), \( f(7)=k \), \( f(9)=26 \)

\( f(5)=25a + 5b + c = 1 \)

\( f(7)=49a + 7b + c = k \)

\( f(9)=81a + 9b + c = 26 \)

Subtract the first equation from the second: \( (49a - 25a)+(7b - 5b)+(c - c)=k - 1 \implies 24a + 2b = k - 1 \)

Subtract the second equation from the third: \( (81a - 49a)+(9b - 7b)+(c - c)=26 - k \implies 32a + 2b = 26 - k \)

Now subtract the first resulting equation from the second: \( (32a - 24a)+(2b - 2b)=(26 - k)-(k - 1) \implies 8a = 27 - 2k \)

We know \( a = 0.75 \), so \( 8\times0.75 = 27 - 2k \implies 6 = 27 - 2k \implies 2k = 21 \implies k = 10.5 \)? Wait, but that's a decimal. Wait, maybe I made a mistake in the interpretation. Wait, the problem says "the average rates of change over equal intervals of size 1 of \( f \) are changing at a constant rate of 1.5". So the average rate of change over \([5,6]\) is \( m \), over \([6,7]\) is \( m + 1.5 \), over \([7,8]\) is \( m + 3 \), over \([8,9]\) is \( m + 4.5 \). Then the total change from 5 to 9 is \( (m)+(m + 1.5)+(m + 3)+(m + 4.5)=4m + 9 \). And the total change is \( 26 - 1 = 25 \), so \( 4m + 9 = 25 \implies 4m = 16 \implies m = 4 \). Then the average rate over \([5,6]\) is 4, so \( f(6)=f(5)+4 = 5 \). Over \([6,7]\) is \( 4 + 1.5 = 5.5 \), so \( f(7)=f(6)+5.5 = 5 + 5.5 = 10.5 \). Over \([7,8]\) is \( 5.5 + 1.5 = 7 \), so \( f(8)=10.5 + 7 = 17.5 \). Over \([8,9]\) is \( 7 + 1.5 = 8.5 \), so \( f(9)=17.5 + 8.5 = 26 \), which matches. So \( k = 10.5 \)? But that's a fraction. Wait, maybe the problem has integer values, so maybe my interpretation is wrong. Wait, maybe the "average rates of change over equal intervals of size 1" are changing at a constant rate, meaning that the difference between the average rates is constant (i.e., the second difference of the function values is constant). For a quadratic function, the second differences are constant. Let's list the x-values: 5, 7, 9. The intervals between x-values are 2 (from 5 to 7) and 2 (from 7 to 9). Wait, maybe the problem is that the average rate of change over [5,7] and [7,9]? Wait, no, the problem says "equal intervals of size 1". So the intervals are [5,6], [6,7], [7,8], [8,9], each of size 1. So the average rate over [5,6] is \( \frac{f(6)-f(5)}{1} \), over [6,7] is \( \frac{f(7)-f(6)}{1} \), etc. The rate of change of these average rates is constant at 1.5, so the difference between consecutive average rates is 1.5. So let's denote:

Let \( f(5) = 1 \), \( f(6) = 1 + d \), \( f(7) = 1 + d + (d + 1.5) = 1 + 2d + 1.5 \), \( f(8) = 1 + 2d + 1.5 + (d + 3) = 1 + 3d + 4.5 \), \( f(9) = 1 + 3d + 4.5 + (d + 4.5) = 1 + 4d + 9 \)

We know \( f(9) = 26 \), so \( 1 + 4d + 9 = 26 \implies 4d + 10 = 26 \implies 4d = 16 \implies d = 4 \)

Then \( f(7) = 1 + 2\times4 + 1.5 = 1 + 8 + 1.5 = 10.5 \). So \( k = 10.5 \), which is \( \frac{21}{2} \). But maybe the problem expects a fractional or decimal answer. Alternatively, maybe I made a mistake in the sign. Wait, if the average rates are changing at a constant rate of 1.5, maybe it's increasing by 1.5 each time, which is what I did. Let's check the total change: from 5 to 6: 4, 6 to 7: 5.5, 7 to 8: 7, 8 to 9: 8.5. Sum: 4 + 5.5 = 9.5, 9.5 + 7 = 16.5, 16.5 + 8.5 = 25. And 26 - 1 = 25, which matches. So \( k = 10.5 \) or \( \frac{21}{2} \).

Wait, but maybe the problem is that the average rate of change over [5,7] and [7,9]? Let's try that. The interval from 5 to 7 is size 2, so average rate is \( \frac{k - 1}{2} \). From 7 to 9 is size 2, average rate is \( \frac{26 - k}{2} \). The rate of change of these average rates (over intervals of size 2) is 1.5? No, the problem says "equal intervals of size 1". So my first interpretation is correct.

Answer:

\( \boxed{10.5} \) (or \( \boxed{\frac{21}{2}} \))