How do you find the slope of the secant lines of #f (x) = x ^2 - 4x + 5 # at (1, 2) and (5, 10)?

Question

Isabella Ackerman · Accepted Answer

Slope of secant is #2#.

First let us confirm, whether #(1,2)# and #(5,10)# lies on #f(x)=x^2-4x+5#. As #1^2-4*1+5=1-4+5=2#, #(1,2)# lies on the curve and as #5^2-4*5+5=25-20+5=10#, #(5,10)# lies on the curve. Now slope of line joining #(x_1,y_1)# and #(x_2,y_2)# is #(y_2-y_1)/(x_2-x_1)# and hence slope of secant is #(10-2)/(5-1)=8/4=2# graph{(x^2-4x+5-y)((x-1)^2+(y-2)^2-0.01)((x-5)^2+(y-10)^2-0.01)(y-2x)=0 [-7.13, 12.87, 0.92, 10.92]}

Emily Ammerman · Answer

undefined To find the slope of the secant line between two points on a function, you use the formula:

$$ \text{Slope} = \frac{{\text{Change in } y}}{{\text{Change in } x}} $$

For the function $ f(x) = x^2 - 4x + 5 $, evaluate the function at the given x-values to find the corresponding y-values. Then, use these points to calculate the slope using the formula mentioned above.

At $ x = 1 $, $ f(1) = 1^2 - 4(1) + 5 = 2 $. At $ x = 5 $, $ f(5) = 5^2 - 4(5) + 5 = 10 $.

So, the points are $ (1, 2) $ and $ (5, 10) $.

The change in $ y $ is $ 10 - 2 = 8 $, and the change in $ x $ is $ 5 - 1 = 4 $.

Therefore, the slope of the secant line is $ \frac{8}{4} = 2 $.