SOLUTION: Find the equation of the tangent line to y=3x^2+0.6x-3 at x=1.5

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Question 887420: Find the equation of the tangent line to y=3x^2+0.6x-3 at x=1.5
Found 2 solutions by Alan3354, nerdybill:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
Find the equation of the tangent line to y=3x^2+0.6x-3 at x=1.5
y=3x^2+0.6x-3
@x = 1.5, y = 4.65 --> tangent at (1.5,4.65)
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y' = 6x + 0.6
@x = 1.5, y' = 9.6
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y - 4.65 = 9.6*(x - 1.5)

Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website!
Find the equation of the tangent line to y=3x^2+0.6x-3 at x=1.5
.
y-value when x=1.5
y=3x^2+0.6x-3
y=3(1.5)^2+0.6(1.5)-3
y=6.75+0.9-3
y=4.65
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Now, you have a point at (1.5, 4.65)
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Next find the first derivative:
y' = 6x + 0.6
slope at x=1.5 is
6(1.5) + 0.6
9.6 (slope)
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Now, using the slope (9.6) and the single point (1.5, 4.65) plug into "point-slope" form:
y - y1 = m(x - x1)
y - 4.65 = 9.6(x - 1.5)
y - 4.65 = 9.6x - 14.4
y = 9.6x - 9.75 (equation of line in "slope-intercept" form)