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Exercise 8.9 · Q1

Q.For the function f(x)=x3+3x2+1f(x) = x^3+3x^2+1 given above, the tangent line at point x=−3x=-3 is drawn using GeoGebra graphing calculator. Draw the tangent line using this application at x=−2x = -2 and x=1x = 1.

Ladakh CbseNCERTSubjective· 2mImportance★★★★★est
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Compute f′(x)f'(x), evaluate slope and point at each given xx, and write the tangent-line equation y−y0=f′(x0)(x−x0)y-y_0=f'(x_0)(x-x_0) — these are the lines to draw in GeoGebra.

Equation of the tangent to y=f(x)y=f(x) at x=x0x=x_0:

y−f(x0)=f′(x0) (x−x0)y - f(x_0) = f'(x_0)\,(x-x_0)

with f(x)=x3+3x2+1  ⟹  f′(x)=3x2+6xf(x)=x^3+3x^2+1 \implies f'(x) = 3x^2+6x.

  1. Differentiate f(x)f(x).

f′(x)=ddx(x3+3x2+1)=3x2+6xf'(x) = \frac{d}{dx}\left(x^3+3x^2+1\right) = 3x^2 + 6x

  1. Tangent at x=−2x=-2. Find f(−2)f(-2) and f′(−2)f'(-2):

f(−2)=(−2)3+3(−2)2+1=−8+12+1=5f(-2) = (-2)^3+3(-2)^2+1 = -8+12+1 = 5

f′(−2)=3(−2)2+6(−2)=12−12=0f'(-2) = 3(-2)^2+6(-2) = 12-12 = 0

Tangent-line equation: y−5=0⋅(x−(−2))y - 5 = 0\cdot(x-(-2)), i.e.

y=5(a horizontal line — the curve has a turning point near here)y = 5 \quad \text{(a horizontal line — the curve has a turning point near here)}

  1. Tangent at x=1x=1. Find f(1)f(1) and f′(1)f'(1):

f(1)=13+3(1)2+1=1+3+1=5f(1) = 1^3+3(1)^2+1 = 1+3+1 = 5

f′(1)=3(1)2+6(1)=3+6=9f'(1) = 3(1)^2+6(1) = 3+6 = 9

Tangent-line equation: y−5=9(x−1)y - 5 = 9(x-1)

y=9x−9+5=9x−4y = 9x - 9 + 5 = 9x - 4

  1. To draw in GeoGebra: type f(x)=x^3+3x^2+1, then Tangent((-2,f(-2)),f) and Tangent((1,f(1)),f) (or directly type y=5 and y=9x-4) to overlay both tangent lines on the curve.

  2. Self-check. At x=−2x=-2: since f′(−2)=0f'(-2)=0, the curve momentarily flattens — matches a local extremum there (in fact ff has a local max at x=−2x=-2, since f′f' changes sign). At x=1x=1: substituting back, the tangent line gives y=9(1)−4=5=f(1)y=9(1)-4=5=f(1). ✓ Point lies on the tangent, as required.

✓Final answer

Tangent at x=−2x=-2: y=5\boxed{y=5}. Tangent at x=1x=1: y=9x−4\boxed{y=9x-4}.

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