(C) UCLES 20239709/21/M/J/2332 A curve has equation y=1+2x2+3lnx.Find the equation of the tangent to the curve at the point (1,32). Give your answer in the form ax+by+c=0, where a,b and c are integers.[5]
Q. (C) UCLES 20239709/21/M/J/2332 A curve has equation y=1+2x2+3lnx.Find the equation of the tangent to the curve at the point (1,32). Give your answer in the form ax+by+c=0, where a,b and c are integers.[5]
Find Derivative: First, we need to find the derivative of the function y=1+2x2+3lnx to determine the slope of the tangent at the given point (1,32).
Use Point-Slope Form: Now, we use the point-slope form of the equation of a line, y−y1=m(x−x1), where m is the slope and (x1,y1) is the point on the line. Here, m=95 and (x1,y1)=(1,32).
More problems from Find equations of tangent lines using limits