Question

f(x)=40+4x+x^2 (e) What is the slope of the chord connecting the points (5, f (5)) and (5 + ∆x, f (5 + ∆x)), when ∆x = 5? Really confused on how to do this and what it means

Answer 1

Answer:

19

Step-by-step explanation:

So we have the function:

$f(x)=40+4x+x^2$

And we want to find the slope of the chord that connects the points (5,f(5)) and (5+Δx, f(5+Δx)).

So first, find the two points.

1) (5,f(5))

Substitute in 5 for x:

$f(x)=40+4x+x^2\\f(5)=40+4(5)+(5)^2$

Square and multiply:

$f(5)=40+20+25$

Add:

$f(5)=85$

So, our first point is (5,85)

2) (5+Δx, f(5+Δx))

Remove the Δx by substituting it with 5. Thus:

$(5+\Delta x,f(5+\Delta x))=(5+5,f(5+5))=(10,f(10))$

Now, substitute in 10 for x. Thus:

$f(10)=40+4(10)+(10)^2$

Square and multiply:

$f(10)=40+40+100$

Add:

$f(10)=180$

So our second point is (10,180).

Now, just find the slope between (5,85) and (10,180) using the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Let (5,85) be x₁ and y₁ and let (10,180) be x₂ and y₂. Substitute:

$m=\frac{180-85}{10-5}$

Subtract:

$m=\frac{95}{5}$

Divide:

$m=19$

So, the slope that connects the chord is 19.

And we are done :)