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In the diagrams below, three points with equally spaced x-coordinates are shown. (0,1) (2,5) mi a. At left, a piecewise linear graph is constructed from the three points. Give the formula for y along the separate pieces in this diagram.

In the diagrams below, three points with equally spaced x-coordinates are shown. (0,1) (2,5) mi a. At left, a piecewise linear graph is constructed from the three points. Give the formula for y along the separate pieces in this diagram.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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HI please do not reject this question even if you just answer part a that would be helpful thank you!!

8. In the diagrams below, three points with equally spaced x-coordinates are shown. (0,1) (2,5) (4,3) to show that a. At left, a piecewise linear graph is constructed from the three points. Give the formula for y along the separate pieces in this diagram. b. At right, the piecewise linear formula gives for the general diagram Yright = Ymiddle + m₂ Ax Yleft=Ymiddle - my Ax C y2 = y1 +m(x2-x₁) +- (x₂- mi Ax Use this to show that the slope m in the diagram is "2+. Hint: m = 2 c. Continue with the general case at right. It is true that three noncollinear points determine a unique parabola. The formula for this parabola is easiest to find as y = y₁ + m(x-x₁) + (x-x₁)², where (x₁, y₁) is the point in the middle (with (xo, yo) to the left and (x2, y2) on the right just to simplify notation). The value of m is from part b). Use (x₂-x₁)² Yright-Yleft Fright-left с m₂ = m +- (Ax) 2 and then use the new formula for m to solve for c. Conclusion: The parabola through three equally spaced points Po, P1, P2 has formula C m2 + mi m₂ - mi y = y₁ +m(x-x₁) +- (x-x₁)² with m = 2 2 Ax Formula (1) is very useful for fitting a parabola to a given set of three points. Note that if the points are collinear, c = 0. and c= d) Use your work in part a) to verify that each of the points in the left-side diagram, (0, 1), (2, 5), and (4, 3), satisfies formula (1) above.
Transcribed Image Text:8. In the diagrams below, three points with equally spaced x-coordinates are shown. (0,1) (2,5) (4,3) to show that a. At left, a piecewise linear graph is constructed from the three points. Give the formula for y along the separate pieces in this diagram. b. At right, the piecewise linear formula gives for the general diagram Yright = Ymiddle + m₂ Ax Yleft=Ymiddle - my Ax C y2 = y1 +m(x2-x₁) +- (x₂- mi Ax Use this to show that the slope m in the diagram is "2+. Hint: m = 2 c. Continue with the general case at right. It is true that three noncollinear points determine a unique parabola. The formula for this parabola is easiest to find as y = y₁ + m(x-x₁) + (x-x₁)², where (x₁, y₁) is the point in the middle (with (xo, yo) to the left and (x2, y2) on the right just to simplify notation). The value of m is from part b). Use (x₂-x₁)² Yright-Yleft Fright-left с m₂ = m +- (Ax) 2 and then use the new formula for m to solve for c. Conclusion: The parabola through three equally spaced points Po, P1, P2 has formula C m2 + mi m₂ - mi y = y₁ +m(x-x₁) +- (x-x₁)² with m = 2 2 Ax Formula (1) is very useful for fitting a parabola to a given set of three points. Note that if the points are collinear, c = 0. and c= d) Use your work in part a) to verify that each of the points in the left-side diagram, (0, 1), (2, 5), and (4, 3), satisfies formula (1) above.
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