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Find the equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen. y = 6x – 5Vx, (1, 1)

Find the equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen. y = 6x – 5Vx, (1, 1)

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter3: Straight Lines And Linear Functions
Section3.1: The Geometry Of Lines
Problem 1SBE: A Ramp A ramp runs from the ground level to a building entrance that is 3 feet high and 12 feet from...
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Step 3 Now since 1-1/2 = 1 we have the following. m = f '(1) -5/2 Submit Skip (you cannot come back)
Transcribed Image Text:Step 3 Now since 1-1/2 = 1 we have the following. m = f '(1) -5/2 Submit Skip (you cannot come back)
Find the equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen. y = 6x - 5/x, (1, 1) Step 1 We can find the equation of a line by using the Point-Slope formula y - Yo = m(x - xo). This means that we need to find the slope m of the line and a point (xo, Yo) on the line. We begin by finding the slope. The line tangent to f(x) at (xo, Yo) will have the slope m = f '(xo). We have y = f(x) = 6x – 5/x. Since 5Vx = 5x1/2, then -1/2 f '(x) = 6 Step 2 Using f '(x) = 6 - we can now find the slope of the tangent line at (1, 1) by finding the following. m = f 1 -1/2 = 6 -
Transcribed Image Text:Find the equation of the tangent line to the curve at the given point. Illustrate by graphing the curve and the tangent line on the same screen. y = 6x - 5/x, (1, 1) Step 1 We can find the equation of a line by using the Point-Slope formula y - Yo = m(x - xo). This means that we need to find the slope m of the line and a point (xo, Yo) on the line. We begin by finding the slope. The line tangent to f(x) at (xo, Yo) will have the slope m = f '(xo). We have y = f(x) = 6x – 5/x. Since 5Vx = 5x1/2, then -1/2 f '(x) = 6 Step 2 Using f '(x) = 6 - we can now find the slope of the tangent line at (1, 1) by finding the following. m = f 1 -1/2 = 6 -
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