It would equal 93.2057mph
The point-slope form of the equation for a line can be written as
... y = m(x -h) +k . . . . . . . for a line with slope m through point (h, k)
Your function gives
... f'(h) = m
... f(h) = k
a) The tangent line is then
... y = 5(x -2) +3
b) The normal line will have a slope that is the negative reciprocal of that of the tangent line.
... y = (-1/5)(x -2) +3
_____
You asked for "an equation." That's what is provided above. Each can be rearranged to whatever form you like.
In standard form, the tangent line's equation is 5x -y = 7. The normal line's equation is x +5y = 17.
Answer:
Step-by-step explanation:
Around 0
Answer:
y=-
x+
Step-by-step explanation:
First, calculate the slope of the line that is perpendicular to the equation of line we are asked to find
m=(y2-y1)/(x2-x1)
=(2-(-4))/(-2-1)
=6/-3
=-2
in this equation the slope is 2, and to find the first equation, use y=mx+b
use the point (1, -4) to find b
-4=(2)(1)+b
-4=2+b
b=-6
the first equation of the line is y=2x-6
to find the x intercept of that line substitute 0 for y
0=2x-6
2x=6
x=3
the slope of a line perpendicular to this would be the opposite reciprocal of the slope which would be equal to -1/2
for the second equation of the line to pass thorugh the x-intercept of the first line, it must pass through (3, 0), so substitute and solve for b
y=mx+b
0=(-1/2)(3)+b
b=3/2
thus the equation of the line that is perpendicular to the line through (1,-4) and (-2, 2) and passes through the x intercept of that line is y=-
x+3/2