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Author Jonathan David Author of Fiction
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Tangent Line and Velocity Problems | Find the Tangent Line to y = x² at x = 3/2

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Sample from The Ultimate Crash Course Series
Get the full bundle with structured lessons and over 1,000 lessons and podcasts via Payhip and theSTEMmajor.com.

The Tangent and Velocity Problems

Point Slope Form Reminder

y-y_0=m(x-x_0)

y=mx+b

Calculus Tangent Line Formula

y-f(a)=f'(a)(x-a)

y_{tangent}=f'(a)(x-a)+f(a)

Find the Tangent Line to

y=x^2 \quad \text{at} \quad x=\frac{3}{2}

Step 1: Identify the Function

f(x)=x^2

x_0=\frac{3}{2}

Step 2: Compute the Function Value

f\left(\frac{3}{2}\right)=\left(\frac{3}{2}\right)^2=\frac{9}{4}

Step 3: Compute the Derivative

f'(x)=2x

f'\left(\frac{3}{2}\right)=2\left(\frac{3}{2}\right)=3

Step 4: Plug Into Tangent Formula

y-f(a)=f'(a)(x-a)

y-\frac{9}{4}=3\left(x-\frac{3}{2}\right)

Step 5: Convert to Slope Intercept Form

y-\frac{9}{4}=3x-\frac{9}{2}

y=3x-\frac{9}{2}+\frac{9}{4}

y=3x-\frac{18}{4}+\frac{9}{4}

y=3x-\frac{9}{4}

Final Tangent Line

y=3x-\frac{9}{4}

Interpretation

The slope of the tangent line equals the instantaneous velocity of the function at the point. Because the derivative of x squared is 2x, the slope at x equals three halves is three. This creates the tangent line that touches the parabola at exactly one point.

Sample from The Ultimate Crash Course Series
This lesson is part of the Ultimate Crash Course bundle. Get the full collection at Payhip and access over 1,000 lessons and podcasts through theSTEMmajor.com.
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