Let f be the function defined by f(x)=(4x-8)/(x^2+5x-14)

write an equation of the lien normal to the graph of f at x=1

how?

the line tangent to the graph at (x,y) has slope f'(x)

f'(x) = -4/(x+7)^2
f'(1) = -4/8^2 = -1/16

The line normal to f thus has slope +16

f(1) = 1/2

So, now we have a point (1,1/2) and a slope: 16

So, the line is

y - 1/2 = 16(x-1)

To find the equation of the line that is normal to the graph of f at x = 1, we first need to find the slope of the tangent line at x = 1.

Step 1: Find the derivative of f(x):
To find the derivative of f(x), we can use the quotient rule.
f'(x) = [(x^2+5x-14)(4) - (4x-8)(2x+5)] / (x^2+5x-14)^2
Simplifying this expression gives f'(x) = [4x^2 + 20x - 56 - 8x^2 - 40x - 40x - 56] / (x^2+5x-14)^2
Simplifying further gives f'(x) = (-4x^2 - 60x - 112) / (x^2 + 5x - 14)^2

Step 2: Find the slope of the tangent line at x = 1:
To find the slope at x = 1, substitute x = 1 into f'(x):
f'(1) = (-4(1)^2 - 60(1) - 112) / (1^2 + 5(1) - 14)^2
f'(1) = (-4 - 60 - 112) / (1 + 5 - 14)^2
f'(1) = (-176) / (-8)^2
f'(1) = -176 / 64
f'(1) = -11/4

Step 3: The slope of the normal line is the negative reciprocal of the slope of the tangent line. So the slope of the normal line is 4/11.

Step 4: Use the point-slope form of a line to write the equation of the line:
The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.
Since we are looking for a line normal to the graph of f at x = 1, we can use the point (1, f(1)) as the point on the line. To find y1, substitute x = 1 into f(x):
f(1) = (4(1) - 8) / (1^2 + 5(1) - 14)
f(1) = (4 - 8) / (1 + 5 - 14)
f(1) = -4 / -8
f(1) = 1/2
So the point on the line is (1, 1/2).

Now we can substitute the slope (4/11) and the point (1, 1/2) into the point-slope form of a line:
y - 1/2 = (4/11)(x - 1)
y - 1/2 = 4/11x - 4/11
y = 4/11x - 4/11 + 1/2
y = 4/11x - 4/11 + 5/11
y = (4x - 3)/11

Therefore, the equation of the line normal to the graph of f at x = 1 is y = (4x - 3)/11.

To find the equation of the line normal to the graph of a function at a given point, we need to follow these steps:

1. Find the derivative of the function f(x).
2. Evaluate the derivative at the given point to obtain the slope of the tangent line.
3. Determine the negative reciprocal of the tangent line's slope to get the slope of the normal line.
4. Use the point-slope form of a line to write the equation of the normal line using the slope from step 3 and the given point.

Let's go through each step for the given function f(x) = (4x - 8) / (x^2 + 5x - 14):

Step 1: Find the derivative of f(x)
The derivative of f(x) is found by applying the quotient rule and simplifying it:

f'(x) = (d/dx)(4x - 8) / (x^2 + 5x - 14)
= (4(x^2 + 5x - 14) - (4x - 8)(2x + 5)) / (x^2 + 5x - 14)^2
= (4x^2 + 20x - 56 - 8x^2 - 20x - 40x - 40) / (x^2 + 5x - 14)^2
= (-4x^2 - 40x - 96) / (x^2 + 5x - 14)^2
= -4(x^2 + 10x + 24) / (x^2 + 5x - 14)^2

Step 2: Evaluate the derivative at x = 1
Substitute x = 1 into the derivative:

f'(1) = -4(1^2 + 10(1) + 24) / (1^2 + 5(1) - 14)^2
= -4(1 + 10 + 24) / (1 + 5 - 14)^2
= -4(35) / (-8)^2
= 140 / 64
= 35/16

The slope of the tangent line at x = 1 is 35/16.

Step 3: Determine the negative reciprocal of the slope
To find the slope of the normal line, we take the negative reciprocal of the tangent line's slope:

Slope of the normal line = -1 / (35/16)
= -16/35

Step 4: Use the point-slope form to write the equation of the normal line
The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.

We are given that the point is x = 1. Let's substitute the values into the equation:

y - y1 = m(x - x1)
y - f(1) = -16/35(x - 1)

Now, let's find the value of f(1) by substituting into the original function:

f(1) = (4(1) - 8) / (1^2 + 5(1) - 14)
= -4 / (-8)
= 1/2

Substituting this back into the equation:

y - 1/2 = -16/35(x - 1)

Therefore, the equation of the line normal to the graph of f(x) at x = 1 is y - 1/2 = -16/35(x - 1).