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MATH-UA 120 Lecture Notes - Lecture 12: Completing The Square, Nth Root, Rational Function

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7 Sep 2016

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New York University

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Domain: all possible x values of an equation. To find the domain, find all solutions of denominator by setting it equal to zero. **the domain is all real numbers except the solutions. To find the zeros, find the solutions of the numerator by setting it equal to zero. ***anything that is not part of your domain should not be in your zeros*** To divide rational expressions, multiply first expression by reciprocal of second. Lcd is the smallest denominator that each denominator divides evenly into. To solve, multiply entire fraction by lcd and simplify. Method 1: simplify numerator and denominator separately and simplify results. Method 2: multiply all fractions by the lcd of all the fractions in the problem, then simplify. Method 1: simplify each side separately, cross multiply, and solve. **reject solutions that make the original equation 0** Method 2: multiply whole equation by lcd, and solve. Three ways to express relations: graph, table, mapping diagram.

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Related Questions

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Using the function y = f(x) obtained in (c), find the .y coordinate corresponding to the value x = -1 / 2 Evaluate both formulas for dy / dx at this point. Why do you think this happens.? Using the explicit derivative formula dy / dx = f (x) obtained in part (c) , find all points on the graph of y = f(x) where the tangent line is horizontal. Then substitute the x and y - coordinates of each such point into the implicit dy / dx formula given in part(b). Again, why does this happen? Again using the explicit derivative formula dy / dx = f(x) from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit dy / dx formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation 2x = y - x2 / x +2y. In effect, points on the line x + 2y = 0 are "cut out" of the graph of the function f Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve for y ? Explain. [ Verification of the implicit derivative given in part (b).] Implicitly differentiate the equation 2x = y - x2 / x = 2y, applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for dy / dx and simplify. Note: Solutions to this problem set are t0 be established by traditional algebraic or arithmetic means. using Maple software to verify each result To download the Maple template file prepared to assist you w ith these exercises, go to https;//my,ltu edu and login, select Calculus 1 and click on Workshop. Saw this file to your laptop hard drive and exit the web Hand-generated solutions are to be fully documented In the workshop report including all steps requiring algebraic or calculus computations - merely listing the answers produce with Maple is not sufficient. Printouts of your Maple verifications are required as part of your workshop report. Consider the equation 2x = y -x2 / x+ 2y. Verify that the points .A(1, -1) and B(-2, 4/3) lie on the graph of this equation. In part (j) of this worksheet you will be asked to verify- the derivative formula by implicitly differentiating the equation given above. Evaluate this derivative at points A and B. Solve for y in the original equation, expressing y as an explicit function of x. (For later reference, we will call this function f(x).) Compute dy / dx from this expression and simplify , then evaluate this explicit version of dy / dx at both A aud B. Considering the results of part (b), are these the values you expected? Why or why not? Evaluate each of the two formulas for dy / dx (both implicit and explicit) at the point C(-1, 0). Can you explain your results?

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Using the function y = f(x) obtained in (c), find the .y coordinate corresponding to the value x = -1 / 2 Evaluate both formulas for dy / dx at this point. Why do you think this happens.? Using the explicit derivative formula dy / dx = f (x) obtained in part (c) , find all points on the graph of y = f(x) where the tangent line is horizontal. Then substitute the x and y - coordinates of each such point into the implicit dy / dx formula given in part(b). Again, why does this happen? Again using the explicit derivative formula dy / dx = f(x) from part (c). find all points on the graph of f where the tangent slope equals 6. Now substitute the x- and y- coordinates of each such point into the implicit dy / dx formula given in part (b) to confirm this slope value. It is clear that no point (x, y) for which x + 2y = 0 will appear on the graph of original equation 2x = y - x2 / x +2y. In effect, points on the line x + 2y = 0 are "cut out" of the graph of the function f Relate this observation to the phenomena encountered earlier in parts (e) and (f). Once again referring to the explicit function f(x) obtained in part (c), where does the graph of f have a vertical asymptote? What happens when you substitute this X-value into the original implicit equation (given at the beginning of this worksheet) and attempt to solve for y ? Explain. [ Verification of the implicit derivative given in part (b).] Implicitly differentiate the equation 2x = y - x2 / x = 2y, applying the quotient role on the right-hand side. Clear the denominator and expand terms, then solve for dy / dx and simplify. Note: Solutions to this problem set are t0 be established by traditional algebraic or arithmetic means. using Maple software to verify each result To download the Maple template file prepared to assist you w ith these exercises, go to https;//my,ltu edu and login, select Calculus 1 and click on Workshop. Saw this file to your laptop hard drive and exit the web Hand-generated solutions are to be fully documented In the workshop report including all steps requiring algebraic or calculus computations - merely listing the answers produce with Maple is not sufficient. Printouts of your Maple verifications are required as part of your workshop report. Consider the equation 2x = y -x2 / x+ 2y. Verify that the points .A(1, -1) and B(-2, 4/3) lie on the graph of this equation. In part (j) of this worksheet you will be asked to verify- the derivative formula by implicitly differentiating the equation given above. Evaluate this derivative at points A and B. Solve for y in the original equation, expressing y as an explicit function of x. (For later reference, we will call this function f(x).) Compute dy / dx from this expression and simplify , then evaluate this explicit version of dy / dx at both A aud B. Considering the results of part (b), are these the values you expected? Why or why not? Evaluate each of the two formulas for dy / dx (both implicit and explicit) at the point C(-1, 0). Can you explain your results?

must understand functions to be able linearize. First let's review what graphs of certain functions looks like. Sketch the shape of each type of y vs. x function below. k is listed as a generic constant of proportionally. Linear y = kx Inverse y = k/x Inverse Square y = k/x^2 Power y = kx^2 You will notice that only the linear function is a straight line. We can easily find the slope of our line by measuring the rise and dividing it by the run of the graph or calculating it using two points. The value of the slope should equal the constant, k, from the equation. Finding k is a bit more challenging in the last three graphs because the slope isn't constant. This should make sense since your graphs aren't linear. So how do you calculate the constant, k? You need to transform the non-linear graph into a linear graph in order to calculate a constant slope. You can accomplish this by transforming one or both of the axes for the graph. The hardest part is figuring out which axes to change and how to change them. The easiest way to accomplish this task is to solve your equation for the constant. Note in the examples from the last page there is only one constant, but this process could be done for other equations with multiple constants. Instead of solving for a single constant, put all of the constants on one side of the equation. When you solve for the constant, the other side of the equation should be in fraction form. This fraction gives the rise and run of the linear graph. Whatever is in the numerator is the vertical axis and the denominator is the horizontal axis. If the equation is not in fraction form, you will need to inverse one or more of the variables to make a fraction. First let's solve each equation to figure out what we should graph. Then look below at the example and complete the last one, a sample AP question, on your own. State what should be graphed in order to produce a linear graph to solve for k. Inverse Graph Vertical Axis: _____ Horizontal Axis: _____ Inverse Square Graph Vertical Axis: _____ Horizontal Axis: _____ Power (Square) Graph Vertical Axis: _____ Horizontal Axis: _____