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What is the equation of the line tangent to #f(x)=7x^2-3x +6 # at #x=1#?

Answer 1

It is #y=11x-1#.

The equation of a line in the plane is

#y=mx+q#

When we have a function #f(x)#, the line tangent to the function in a specific point has the #m# that is the derivative of the function in that point. That is #m=(df(p_0))/dx# where #p_0# is the desired point. So first of all we have to calculate the derivative

#(df)/dx=14x-3#

then, for #x=1#, we have

#m=(df(1))/dx=11#.

Our line is then

#y=11x+q#. We still miss #q#, but it is enough to know that the line passes for one point to evaluate it. We know that the line is tangent to the function in #x=1#, so we are sure that our line will pass from the point

#x=1, y=f(1)=7*1^2-3*1+6=10#. We substitute this point in the equation of the line

#10=11*1+q# obtaining #q=-1#. Then the equation of the line is

#y=11x-1#.

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Answer 2

Evelyn Agard

The equation of the line tangent to f(x)=7x^2-3x+6 at x=1 is y=8x+10.

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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.