in the xy-plane the line x+y=k where k is a constant is tangent to the graph of y=x2+3x+1 what is the value of k?

Question

faldduE · Accepted Answer

First we rewrite given of a line to obtain y=-x+k. Thus, slope of the line  is m=-1Now the idea is to to use derivative of the function y=x2+3x+1. Derivative at some point represents slope of the tangent line at that point. This means that we have to find a point with deriavative -1.Derivatives of the functions:Find values of for which y′=−1−1=2x+32x=−4x=−2Second coordinate of that point is:y(−2)=(−2)2+3(−2)+1=4−6+1=−1Thus, mentioned line x+y=k passes through point (-2,-1). Now, it is easy to find k using substitution:k=x+yk=−2−1k=−3

fudzisako · Accepted Answer

To determine the value of k for which the line x+y=k is tangent to the graph of y=x2+3x+1 in the xy-plane, we need to find the point of tangency.Let&#039;s first express the equation of the line as y=k−x.Next, we&#039;ll find the derivative of the quadratic function y=x2+3x+1 with respect to x. Taking the derivative, we have dydx=2x+3.The line is tangent to the graph at the point of tangency, so the slope of the line should be equal to the derivative of the function at that point. Therefore, we can equate the slopes and solve for x.2x+3=−1Simplifying the equation, we have 2x=−4, which gives us x=−2.Substituting this value of x into the equation of the line, we find:y=k−(−2)y=k+2Now, we can substitute the values of x and y into the equation of the quadratic function to obtain:k+2=(−2)2+3(−2)+1k+2=4−6+1k+2=−1k=−3Hence, the value of k for which the line x+y=k is tangent to the graph of y=x2+3x+1 is k=−3.

Jazz Frenia · Accepted Answer

Answer:k=−2Explanation:Let&#039;s begin by finding the equation of the tangent line. Since the line is tangent to the curve, it will have the same slope as the curve at the point of tangency. The slope of the curve y=x2+3x+1 can be found by taking the derivative with respect to x:dydx=ddx(x2+3x+1)=2x+3.Next, we need to find the x-coordinate of the point of tangency. We can do this by equating the slope of the curve to the slope of the line:2x+3=1.Solving this equation for x, we get:2x=−2,x=−1.Now, let&#039;s substitute this x-coordinate back into the equation of the curve to find the corresponding y-coordinate:y=(−1)2+3(−1)+1=1−3+1=−1.Therefore, the point of tangency is (−1,−1).Since the line x+y=k passes through this point, we can substitute x=−1 and y=−1 into the equation of the line to solve for k:−1+(−1)=k,−2=k.Hence, the value of k is −2.

Andre BalkonE · Accepted Answer

First, let&#039;s find the derivative of the quadratic function y=x2+3x+1 with respect to x. We have:dydx=ddx(x2+3x+1)=2x+3Next, we know that the line x+y=k is tangent to the graph of y=x2+3x+1. The slope of the tangent line is equal to the derivative of the quadratic function at the point of tangency. So we set the derivative equal to the slope of the line:2x+3=1Simplifying this equation, we get:2x=−2⟹x=−1Now, substituting this value of x back into the quadratic function, we can find the corresponding y-coordinate:y=(−1)2+3(−1)+1=−1Therefore, the point of tangency is (−1,−1).Since the line x+y=k is tangent to the graph at this point, we can substitute the coordinates of the point into the equation of the line to find the value of k:(−1)+(−1)=k⟹k=−2Hence, the value of k is −2.

in the xy-plane the line x+y=k where k is a constant is tangent to the graph of

Answered question

Answer & Explanation

New Questions in Differential Calculus