The slope of y = 3x - 4 on the interval [2, 5] is 3 and the slope of y = 2x^2-4x - 2 on the interval [2, 5] is 10
How to determine the slope?
The interval is given as:
x = 2 to x = 5
The slope is calculated as:
[tex]m = \frac{y_2 -y_1}{x_2-x_1}[/tex]
16. y = 3x - 4
Substitute 2 and 5 for x
y = 3*2 - 4 = 2
y = 3*5 - 4 = 11
So, we have:
[tex]m = \frac{11 - 2}{5 - 2}[/tex]
[tex]m = \frac{9}{3}[/tex]
Divide
m = 3
Hence, the slope of y = 3x - 4 on the interval [2, 5] is 3
17. y = 2x^2-4x - 2
Substitute 2 and 5 for x
y = 2 * 2^2 - 4 * 2 - 2 = -2
y = 2 * 5^2 - 4 * 5 - 2 = 28
So, we have:
[tex]m = \frac{28 + 2}{5 - 2}[/tex]
[tex]m = \frac{30}{3}[/tex]
Divide
m = 10
Hence, the slope of y = 2x^2-4x - 2 on the interval [2, 5] is 10
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