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What is the equation of the line tangent to # f(x)=5x^2-3x+2 # at # x=3#?

Answer 1

Ezekiel Alexander

y = 27x -43

The equation can be expressed as y = mx + c in slope-intercept form.

where c is the y-intercept and m is the gradient.

By evaluating f(3), one can determine that f'(3) equals m and c.

#f(x)=5x^2-3x+2#

#rArr f'(x)=10x-3#

in addition to f'(3) = 10(3) - 3 = 27 = m (tangent gradient).

Thus, the partial equation is y = 27x + c.

now #f(3)=5(3)^2-3(3)+2=38#

Because of this, (3,38) is a point on the tangent, and we can find the value of c by substituting this into the partial equation.

38 = 81 + c → c = - 43 for x = 3 and y = 38.

Thus the equation of tangent is #color(red)(|bar(ul(color(white)(a/a)color(black)(y=27x-43)color(white)(a/a)|)))#

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Answer 2

Emily Ammerman

The equation of the line tangent to f(x)=5x^2-3x+2 at x=3 is y=33x-28.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.