Tangent lines are used to approximate a curve near a specific point. In calculus, the slope of the tangent line comes from the derivative evaluated at the point of tangency. Once you have the slope and a point, you can use point-slope form to write the equation of the tangent line.
Notes
Questions
\(\textbf{1)}\) Find the equation of the tangent line to the curve \( f(x)=x^3+3x^2-x \) at the point \( (2,18) \)
The answer is \( y=23x-28 \)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^3+3x^2-x\)
\(\,\,\,\,\,\,f'(x)=3x^2+6x-1\)
\(\,\,\,\,\,\,f'(2)=3(2)^2+6(2)-1\)
\(\,\,\,\,\,\,f'(2)=12+12-1\)
\(\,\,\,\,\,\,f'(2)=23\)
\(\,\,\,\,\,\,\text{Slope at } (2,18) \text{ is }m=23\)
\(\,\,\,\text{Step 2: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\,\text{Point Slope Formula: }y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\,y-18=23(x-2)\)
\(\,\,\,\,\,\,y-18=23x-46\)
\(\,\,\,\,\,\,\text{The answer is } y=23x-28 \)
\(\textbf{2)}\) Find the equation of the tangent line to the curve \( f(x)=x\sqrt{x} \) at the point \((4,8) \)
The answer is \( y=3x-4 \)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,\,\,\, f(x)=x\sqrt{x} \)
\(\,\,\,\,\,\,\,\,\, f(x)=x^1 \cdot x^{1/2} \)
\(\,\,\,\,\,\,\,\,\, f(x)=x^{1+1/2} \)
\(\,\,\,\,\,\,\,\,\, f(x)=x^{3/2} \)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{3}{2}x^{\left(3/2 – 1\right)} \)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{3}{2}x^{1/2} \)
\(\,\,\,\,\,\,\,\,\, f'(4)=\frac{3}{2}(4)^{1/2} \)
\(\,\,\,\,\,\,\,\,\, f'(4)=\frac{3}{2}(2) \)
\(\,\,\,\,\,\,\,\,\, f'(4)=3 \)
\(\,\,\,\,\,\,\,\,\,\text{Slope at } (4,8) \text{ is }m=3\)
\(\,\,\,\text{Step 2: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\,\,\,\,\text{Point Slope Formula: }y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\,\,\,\,y-8=3(x-4)\)
\(\,\,\,\,\,\,\,\,\,y-8=3x-12\)
\(\,\,\,\,\,\,\,\,\,y=3x-4\)
\(\,\,\,\)The answer is \( y=3x-4 \)
\(\textbf{3)}\) Find the equation of the tangent line to the curve \( f(x)=x^2-6x+4 \) and parallel to the line \(y=-4x+9 \)
The answer is \( y=-4x+3 \)
\(\,\,\,\text{Step 1: Parallel lines have the same slope.}\)
\(\,\,\,\,\,\,\,\,\, \text{The slope of the line } y=-4x+9 \text{ is } m=-4\)
\(\,\,\,\text{Step 2: Take the derivative of } f(x)=x^2-6x+4\)
\(\,\,\,\,\,\,\,\,\, f'(x)=2x-6 \)
\(\,\,\,\,\,\,\,\,\, f'(x)=-4 \)
\(\,\,\,\,\,\,\,\,\, 2x-6=-4 \)
\(\,\,\,\,\,\,\,\,\, x=1 \)
\(\,\,\,\text{Step 3: Find the y-coordinate at } x=1 \)
\(\,\,\,\,\,\,\,\,\, f(1)=1^2-6(1)+4=-1\)
\(\,\,\,\text{Step 4: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\,\,\,\, y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\,\,\,\, y-(-1)=-4(x-1)\)
\(\,\,\,\,\,\,\,\,\, y+1=-4x+4\)
\(\,\,\,\,\,\,\,\,\, y=-4x+3\)
\(\,\,\,\,\,\,\,\,\,\text{The answer is } y=-4x+3\)
\(\textbf{4)}\) Find the equation of the tangent line to the curve \( f(x)=\frac{e^x}{x} \) at the point \( (1,e) \)
The answer is \( y=e \)
\(\,\,\,\text{Step 1: Find the derivative using the quotient rule.}\)
\(\,\,\,\,\,\,\,\,\, f(x)=\frac{e^x}{x} \)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{x(e^x)-(e^x)(1)}{x^2}\)
\(\,\,\,\,\,\,\,\,\, f'(x)=\frac{xe^x-e^x}{x^2}\)
\(\,\,\,\,\,\,\,\,\, f'(1)=\frac{1(e^1)-e^1}{1^2}\)
\(\,\,\,\,\,\,\,\,\, f'(1)=\frac{e-e}{1}=0 \)
\(\,\,\,\text{Step 2: Use the slope } m=0 \text{ and the point } (1,e) \)
\(\,\,\,\,\,\,\,\,\, y-e=0(x-1)\)
\(\,\,\,\,\,\,\,\,\, y=e \)
\(\,\,\,\,\,\,\,\,\,\text{The answer is } y=e \)
\(\textbf{5)}\) Find the equation of the tangent line to the curve \( f(x)=5e^x \cos{x} \) at the point \( (0,5) \)
The answer is \( y=5x+5 \)
\(\,\,\,\text{Step 1: Use the product rule for the derivative.}\)
\(\,\,\,\,\,\, f(x)=5e^x \cos{x} \)
\(\,\,\,\,\,\, f'(x)=5e^x \cos{x}-5e^x \sin{x} \)
\(\,\,\,\,\,\, f'(0)=5e^0 \cos{0}-5e^0 \sin{0} \)
\(\,\,\,\,\,\, f'(0)=5(1)(1)-5(1)(0)=5 \)
\(\,\,\,\text{Step 2: Use the slope } m=5 \text{ and the point } (0,5) \)
\(\,\,\,\,\,\, y-5=5(x-0)\)
\(\,\,\,\,\,\, y-5=5x\)
\(\,\,\,\,\,\, y=5x+5 \)
\(\,\,\,\,\,\,\text{The answer is } y=5x+5 \)
\(\textbf{6)}\) Find the equation of the tangent line to the curve \( f(x)=5\sqrt{x}(x+4) \) at the point \( (4,80) \)
The answer is \( y=20x\)
\(\,\,\,\text{Step 1: Distribute } 5x^{1/2} \text{ across } (x+4).\)
\(\,\,\,\,\,\, f(x)=5x^{1/2}(x)+5x^{1/2}(4)\)
\(\,\,\,\,\,\, f(x)=5x^{3/2}+20x^{1/2}\)
\(\,\,\,\text{Step 2: Differentiate using the power rule.}\)
\(\,\,\,\,\,\, f'(x)=\frac{15}{2}x^{1/2}+10x^{-1/2}\)
\(\,\,\,\text{Step 3: Evaluate } f'(x) \text{ at } x=4.\)
\(\,\,\,\,\,\, f'(4)=\frac{15}{2}(4)^{1/2}+10(4)^{-1/2}\)
\(\,\,\,\,\,\, f'(4)=\frac{15}{2}(2)+10(1/2)\)
\(\,\,\,\,\,\, f'(4)=15+5\)
\(\,\,\,\,\,\, f'(4)=20\)
\(\,\,\,\,\,\, \text{Slope at } (4,80) \text{ is } m=20\)
\(\,\,\,\text{Step 4: Use the point and slope to find the equation of the line.}\)
\(\,\,\,\,\,\, y-y_1=m(x-x_1)\)
\(\,\,\,\,\,\, y-80=20(x-4)\)
\(\,\,\,\,\,\, y-80=20x-80\)
\(\,\,\,\,\,\, y=20x\)
\(\,\,\,\,\,\, \text{The answer is } y=20x\)
\(\textbf{7)}\) Find the equation of the tangent line to the curve \(f(x)=x^2+2x\) at the point \((1,3)\)
The answer is \(y=4x-1\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^2+2x\)
\(\,\,\,\,\,\,f'(x)=2x+2\)
\(\,\,\,\,\,\,f'(1)=2(1)+2=4\)
\(\,\,\,\,\,\,\text{Slope at }(1,3)\text{ is }m=4\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-3=4(x-1)\)
\(\,\,\,\,\,\,y-3=4x-4\)
\(\,\,\,\,\,\,y=4x-1\)
\(\,\,\,\)The answer is \(y=4x-1\)
\(\textbf{8)}\) Find the equation of the tangent line to the curve \(f(x)=x^3-2x\) at the point \((2,4)\)
The answer is \(y=10x-16\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^3-2x\)
\(\,\,\,\,\,\,f'(x)=3x^2-2\)
\(\,\,\,\,\,\,f'(2)=3(2)^2-2=10\)
\(\,\,\,\,\,\,\text{Slope at }(2,4)\text{ is }m=10\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-4=10(x-2)\)
\(\,\,\,\,\,\,y-4=10x-20\)
\(\,\,\,\,\,\,y=10x-16\)
\(\,\,\,\)The answer is \(y=10x-16\)
\(\textbf{9)}\) Find the equation of the tangent line to the curve \(f(x)=\sqrt{x+5}\) at the point \((4,3)\)
The answer is \(y=\frac{1}{6}x+\frac{7}{3}\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\sqrt{x+5}\)
\(\,\,\,\,\,\,f(x)=(x+5)^{1/2}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2}(x+5)^{-1/2}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x+5}}\)
\(\,\,\,\,\,\,f'(4)=\frac{1}{2\sqrt{9}}=\frac{1}{6}\)
\(\,\,\,\,\,\,\text{Slope at }(4,3)\text{ is }m=\frac{1}{6}\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-3=\frac{1}{6}(x-4)\)
\(\,\,\,\,\,\,y=\frac{1}{6}x-\frac{4}{6}+3\)
\(\,\,\,\,\,\,y=\frac{1}{6}x+\frac{7}{3}\)
\(\,\,\,\)The answer is \(y=\frac{1}{6}x+\frac{7}{3}\)
\(\textbf{10)}\) Find the equation of the tangent line to the curve \(f(x)=\ln(x+1)\) at the point \((0,0)\)
The answer is \(y=x\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\ln(x+1)\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{x+1}\)
\(\,\,\,\,\,\,f'(0)=\frac{1}{0+1}=1\)
\(\,\,\,\,\,\,\text{Slope at }(0,0)\text{ is }m=1\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-0=1(x-0)\)
\(\,\,\,\,\,\,y=x\)
\(\,\,\,\)The answer is \(y=x\)
\(\textbf{11)}\) Find the equation of the tangent line to the curve \(f(x)=\sin(2x)\) at the point \((0,0)\)
The answer is \(y=2x\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\sin(2x)\)
\(\,\,\,\,\,\,f'(x)=2\cos(2x)\)
\(\,\,\,\,\,\,f'(0)=2\cos(0)=2\)
\(\,\,\,\,\,\,\text{Slope at }(0,0)\text{ is }m=2\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-0=2(x-0)\)
\(\,\,\,\,\,\,y=2x\)
\(\,\,\,\)The answer is \(y=2x\)
\(\textbf{12)}\) Find the equation of the tangent line to the curve \(f(x)=\cos(x)\) at the point \((0,1)\)
The answer is \(y=1\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\cos(x)\)
\(\,\,\,\,\,\,f'(x)=-\sin(x)\)
\(\,\,\,\,\,\,f'(0)=-\sin(0)=0\)
\(\,\,\,\,\,\,\text{Slope at }(0,1)\text{ is }m=0\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-1=0(x-0)\)
\(\,\,\,\,\,\,y=1\)
\(\,\,\,\)The answer is \(y=1\)
\(\textbf{13)}\) Find the equation of the tangent line to the curve \(f(x)=e^{-x}\) at the point \((0,1)\)
The answer is \(y=-x+1\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=e^{-x}\)
\(\,\,\,\,\,\,f'(x)=-e^{-x}\)
\(\,\,\,\,\,\,f'(0)=-e^0=-1\)
\(\,\,\,\,\,\,\text{Slope at }(0,1)\text{ is }m=-1\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-1=-1(x-0)\)
\(\,\,\,\,\,\,y=-x+1\)
\(\,\,\,\)The answer is \(y=-x+1\)
\(\textbf{14)}\) Find the equation of the tangent line to the curve \(f(x)=x^2-4x+6\) at the point \((3,3)\)
The answer is \(y=2x-3\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^2-4x+6\)
\(\,\,\,\,\,\,f'(x)=2x-4\)
\(\,\,\,\,\,\,f'(3)=2(3)-4=2\)
\(\,\,\,\,\,\,\text{Slope at }(3,3)\text{ is }m=2\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-3=2(x-3)\)
\(\,\,\,\,\,\,y-3=2x-6\)
\(\,\,\,\,\,\,y=2x-3\)
\(\,\,\,\)The answer is \(y=2x-3\)
\(\textbf{15)}\) Find the equation of the tangent line to the curve \(f(x)=\frac{x+1}{x}\) at the point \((1,2)\)
The answer is \(y=-x+3\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\frac{x+1}{x}\)
\(\,\,\,\,\,\,f(x)=1+\frac{1}{x}\)
\(\,\,\,\,\,\,f'(x)=-\frac{1}{x^2}\)
\(\,\,\,\,\,\,f'(1)=-\frac{1}{1^2}=-1\)
\(\,\,\,\,\,\,\text{Slope at }(1,2)\text{ is }m=-1\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-2=-1(x-1)\)
\(\,\,\,\,\,\,y=-x+3\)
\(\,\,\,\)The answer is \(y=-x+3\)
\(\textbf{16)}\) Find the equation of the tangent line to the curve \(f(x)=x^3+x\) at the point \((1,2)\)
The answer is \(y=4x-2\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^3+x\)
\(\,\,\,\,\,\,f'(x)=3x^2+1\)
\(\,\,\,\,\,\,f'(1)=3(1)^2+1=4\)
\(\,\,\,\,\,\,\text{Slope at }(1,2)\text{ is }m=4\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-2=4(x-1)\)
\(\,\,\,\,\,\,y-2=4x-4\)
\(\,\,\,\,\,\,y=4x-2\)
\(\,\,\,\)The answer is \(y=4x-2\)
\(\textbf{17)}\) Find the equation of the tangent line to the curve \(f(x)=\tan(x)\) at the point \((0,0)\)
The answer is \(y=x\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\tan(x)\)
\(\,\,\,\,\,\,f'(x)=\sec^2(x)\)
\(\,\,\,\,\,\,f'(0)=\sec^2(0)=1\)
\(\,\,\,\,\,\,\text{Slope at }(0,0)\text{ is }m=1\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-0=1(x-0)\)
\(\,\,\,\,\,\,y=x\)
\(\,\,\,\)The answer is \(y=x\)
\(\textbf{18)}\) Find the equation of the tangent line to the curve \(f(x)=x^4\) at the point \((1,1)\)
The answer is \(y=4x-3\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^4\)
\(\,\,\,\,\,\,f'(x)=4x^3\)
\(\,\,\,\,\,\,f'(1)=4(1)^3=4\)
\(\,\,\,\,\,\,\text{Slope at }(1,1)\text{ is }m=4\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-1=4(x-1)\)
\(\,\,\,\,\,\,y-1=4x-4\)
\(\,\,\,\,\,\,y=4x-3\)
\(\,\,\,\)The answer is \(y=4x-3\)
\(\textbf{19)}\) Find the equation of the tangent line to the curve \(f(x)=x^2+3x\) at the point \((1,4)\)
The answer is \(y=5x-1\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=x^2+3x\)
\(\,\,\,\,\,\,f'(x)=2x+3\)
\(\,\,\,\,\,\,f'(1)=2(1)+3=5\)
\(\,\,\,\,\,\,\text{Slope at }(1,4)\text{ is }m=5\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-4=5(x-1)\)
\(\,\,\,\,\,\,y-4=5x-5\)
\(\,\,\,\,\,\,y=5x-1\)
\(\,\,\,\)The answer is \(y=5x-1\)
\(\textbf{20)}\) Find the equation of the tangent line to the curve \(f(x)=\sqrt{x^2+3}\) at the point \((1,2)\)
The answer is \(y=\frac{1}{2}x+\frac{3}{2}\)
\(\,\,\,\text{Step 1: Find the slope with the derivative.}\)
\(\,\,\,\,\,\,f(x)=\sqrt{x^2+3}\)
\(\,\,\,\,\,\,f(x)=(x^2+3)^{1/2}\)
\(\,\,\,\,\,\,f'(x)=\frac{1}{2}(x^2+3)^{-1/2}(2x)\)
\(\,\,\,\,\,\,f'(x)=\frac{x}{\sqrt{x^2+3}}\)
\(\,\,\,\,\,\,f'(1)=\frac{1}{\sqrt{1^2+3}}=\frac{1}{2}\)
\(\,\,\,\,\,\,\text{Slope at }(1,2)\text{ is }m=\frac{1}{2}\)
\(\,\,\,\text{Step 2: Use point-slope form.}\)
\(\,\,\,\,\,\,y-2=\frac{1}{2}(x-1)\)
\(\,\,\,\,\,\,y=\frac{1}{2}x-\frac{1}{2}+2\)
\(\,\,\,\,\,\,y=\frac{1}{2}x+\frac{3}{2}\)
\(\,\,\,\)The answer is \(y=\frac{1}{2}x+\frac{3}{2}\)
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