If f(x) is a parabola with vertex (2,−1) whose equation isf(x) =ax2+bx+c, what are the values of a, b, c if the graph off inters
ects the y-axis at the point (0,3).
1 answer:
Answer:
(a, b, c) = (1, -4, 3)
Step-by-step explanation:
In vertex form, the equation is ...
... f(x) = a(x -2)² -1
This has a y-intercept of ...
... f(0) = a(0 -2)² -1 = 4a -1
We want that value to be 3, so we can find "a" as ...
... 4a -1 = 3 . . . . . set f(0) = 3
... 4a = 4 . . . . . . . add 1
... a = 1 . . . . . . . . divide by 4
Then ...
... f(x) = (x -2)² -1 = x² -4x +3
The coefficients are a=1, b=-4, c=3.
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The expression of integral as a limit of Riemann sums of given integral is 4
∑
from i=1 to i=n.
Given an integral .
We are required to express the integral as a limit of Riemann sums.
An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.
A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.
Using Riemann sums, we have :
=
∑f(a+iΔx)Δx ,here Δx=(b-a)/n
=f(x)=
⇒Δx=(5-1)/n=4/n
f(a+iΔx)=f(1+4i/n)
f(1+4i/n)=
∑f(a+iΔx)Δx=
∑
=4∑
Hence the expression of integral as a limit of Riemann sums of given integral is 4
∑
from i=1 to i=n.
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