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37. Calculus The graph of a parabola passes through the points (0, 1) and (3, 2) and has a horizontal tangent line at (,). Find an equation for the parabola and sketch its graph.

37. Calculus The graph of a parabola passes through the points (0, 1) and (3, 2) and has a horizontal tangent line at (,). Find an equation for the parabola and sketch its graph.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.3: Hyperbolas
Problem 46E
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I need help with this Linear Algebra problem, I also put an example, I suspect this is somewhat whatthey want  in this question, but I am nor sure and need help starting it. How should I approach this problem?

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37. Calculus The graph of a parabola passes through the points (0, 1) and (2, 2) and has a horizontal tangent line at (2). Find an equation for the parabola and sketch its graph.
Transcribed Image Text:37. Calculus The graph of a parabola passes through the points (0, 1) and (2, 2) and has a horizontal tangent line at (2). Find an equation for the parabola and sketch its graph.
(-2, 3) 10- 8- (-1,5) 3 Figure 1.5 4- (0, 1) (2, 10) (1,4) EXAMPLE 2 Polynomial Curve Fitting See LarsonLinearAlgebra.com for an interactive version of this type of example. Find a polynomial that fits the points (-2, 3), (-1, 5), (0, 1), (1, 4), and (2, 10). SOLUTION You are given five points, so choose a fourth-degree polynomial function p(x) = a₁ + a₁x + α₂x² + α3x³ + à¸xª. Substituting the points into p(x) produces the system of linear equations shown below. ao - 2a₁ +4a₂-8a3 + 16a4 3 ao- a₁ + a₂ - az + a₁ = 5 ao 1 ao + a₁ + a₂ + az + a4 = 4 a + 2a₁ + 4a₂ + 8a3 + 16a4 = 10 = = The solution of these equations is a₁ = 1, a₁ = -√, A₂ = 24, az = ³/1, A4 = - =-12-17 Figure 1.5 shows the graph of p. which means the polynomial function is p(x) = 1 − ²x + 101x² + x³ – 17x4. -
Transcribed Image Text:(-2, 3) 10- 8- (-1,5) 3 Figure 1.5 4- (0, 1) (2, 10) (1,4) EXAMPLE 2 Polynomial Curve Fitting See LarsonLinearAlgebra.com for an interactive version of this type of example. Find a polynomial that fits the points (-2, 3), (-1, 5), (0, 1), (1, 4), and (2, 10). SOLUTION You are given five points, so choose a fourth-degree polynomial function p(x) = a₁ + a₁x + α₂x² + α3x³ + à¸xª. Substituting the points into p(x) produces the system of linear equations shown below. ao - 2a₁ +4a₂-8a3 + 16a4 3 ao- a₁ + a₂ - az + a₁ = 5 ao 1 ao + a₁ + a₂ + az + a4 = 4 a + 2a₁ + 4a₂ + 8a3 + 16a4 = 10 = = The solution of these equations is a₁ = 1, a₁ = -√, A₂ = 24, az = ³/1, A4 = - =-12-17 Figure 1.5 shows the graph of p. which means the polynomial function is p(x) = 1 − ²x + 101x² + x³ – 17x4. -
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