The tangent line to a curve is the one that coincides with the curve at a point and with the same derivative, that is, the same degree of variation.
We have then:
y = 5x-x²
Deriving:
y '= 5-2x
In point (1, 4)
The slope is:
y (1) '= 5-2 * (1)
y (1) '= 3
The equation of the line will be:
y-f (a) = f '(a) (x-a)
We have then:
y-4 = 3 (x-1)
Rewriting:
y = 3x-3 + 4
y = 3x + 1
Answer:
the tangent line to the parabola at the point (1, 4) is
y = 3x + 1
the slope m is
m = 3
Answer:
24 ft.
Step-by-step explanation:
10x2=20. 20+4=24
First, let's use the given information to determine the function's amplitude, midline, and period.
Then, we should determine whether to use a sine or a cosine function, based on the point where x=0.
Finally, we should determine the parameters of the function's formula by considering all the above.
Determining the amplitude, midline, and period
The midline intersection is at y=5 so this is the midline.
The maximum point is 1 unit above the midline, so the amplitude is 1.
The maximum point is π units to the right of the midline intersection, so the period is 4 * π.
Determining the type of function to use
Since the graph intersects its midline at x=0, we should use thesine function and not the cosine function.
This means there's no horizontal shift, so the function is of the form -
a sin(bx)+d
Since the midline intersection at x=0 is followed by a maximumpoint, we know that a > 0.
The amplitude is 1, so |a| = 1. Since a >0 we can conclude that a=1.
The midline is y=5, so d=5.
The period is 4π so b = 2π / 4π = 1/2 simplified.
f(x)1 sin 1/3x+5 = Solution
Answer:graph 3
Step-by-step explanation:
