Find the slope of the tangent line to the parabola y=x^2+2x at the point (−6,24).
Added by Darren M.
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The given equation is \( y = x^2 + 2x \). Show more…
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EXAMPLE 3 Find the slope of the tangent line to the parabola y = x^2 at the point (2, 4) by analyzing the slopes of secant lines through (2, 4). Write an equation for the tangent line to the parabola at this point. Solution We begin with a secant line through P(2, 4) and a nearby point Q(2 + h, (2 + h)^2). We then write an expression for the slope of the secant line PQ and investigate what happens to the slope as Q approaches P along the curve: Secant line slope = Δy/Δx = ((2 + h)^2 - 2^2)/h = (h^2 + 4h + 4 - 4)/h = (h^2 + 4h)/h = h + 4. If h > 0, then Q lies above and to the right of P, as in Figure 2.4. If h < 0, then Q lies to the left of P (not shown). In either case, as Q approaches P along the curve, h approaches zero and the secant line slope h + 4 approaches 4. We take 4 to be the parabola's slope at P. y = x^2 Secant line slope is ((2 + h)^2 - 4)/h = h + 4. Q(2 + h, (2 + h)^2) Tangent line slope = 4 Δy = (2 + h)^2 - 4 P(2, 4) Δx = h FIGURE 2.4 Finding the slope of the parabola y = x^2 at the point P(2, 4) as the limit of secant line slopes (Example 3).
Donna D.
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