Find the derivative of the function y=2x2−13x+5 and use it to find an equation of the line tangent to the curve at x=3.

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Esther Phillips · Accepted Answer

dydx=limh⇒0y(x+h)−y(x)h =limh⇒0[2(x+h)2−13(x+h)+5]−(2x2−13x+5)h =limh⇒0[2x2+4xh+2h2−13x−13h+5]−2x2+13x−5}{h} =limh⇒0[2x2+4xh+2h2−13x−13h+5−2x2+13x−5}h =limh⇒04xh+2h2−13hh =limh⇒0(4x+2h−13) =4x−13 Step 2 At x=3,dydx=4(3)−13=−1 so the tangent line has slope m=−1 passes through (3,y(3))=(3,−16) y−y1=m(x−x1) y+16=−1(x−3) y=−x−13

Travis Hicks · Accepted Answer

y=2x2−13x+5  dydx=4x−13  At x=3,dydx=4(3)−13=12−13=−1  and y=2(3)2−13(3)+5=2(9)−39+5=18−34=−16  The tangent line at x=3, passes through the point (3, -16) and has a slope of -1  y−(−16)=−1(x−3)  y+16=−x+3  y=−x−13 is the required equation.

nick1337 · Accepted Answer

Since the derivative of a quadratic function is a linear function, then slope can take on ANY real value, including -1.y=2x2−13x+5y′=4x−13=−14x=12x=3y=2(3)2−13(3)+5=18−39+5=−16Point of tangency: (3,-16)Equation of tangent line:y+16=−1(x−3)y=−x−13

Find the derivative of the function y = \frac{2x}{2} -

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