Tangent Line Find the equation(s) of the tangent line(s) to the graph of the curve y=x^3-9 x thr

Tangent Line Find the equation(s) of the tangent line(s) to...

Key Concepts

External Point Condition

When a tangent line is required to pass through a point that is not on the curve, an extra condition is imposed. This involves finding the point(s) of tangency on the curve where the tangent line, determined by the derivative, also passes through the given external point. This leads to an equation (or system of equations) that must be solved to determine all valid points of tangency, ensuring that the line not only touches the curve but also goes through the specified external point.

Point-Slope Form

The point-slope form of a line is a way to express the equation of a line when its slope and at least one point on the line are known. This form is widely used when writing the equation of a tangent line because once the slope (from the derivative) and the point of tangency are identified, the line's equation can be immediately written down.

Derivative

The derivative of a function at a specific point gives the instantaneous rate of change of the function, which is geometrically interpreted as the slope of the tangent line at that point. Computing the derivative is essential in determining how steep the curve is and forms the basis for finding the equation of the tangent line.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point and has the same slope as the curve does there. This concept is fundamental in calculus and is used to locally approximate the curve by a line, providing insight into the behavior of the function near that point.

Transcript

00:01 We have given a function of graph y is equal to x q minus 9x we will compute the slope of the equation of the tangent line which is passing through 1.

00:19 Minus 9 but this is not on the graph so first we will compute the derivative of y with respect to x here so the derivative of y with respect to x is equal to 3x squared minus x.

00:33 Using the power rule of derivative as you know the derivative of y with respect to x gives the slope of the tangent line so we can write m is equal to 3x square minus 9 as you know the slope of the tangent line is y minus y 1 divided by x minus x 1 is equal to 3 x squared minus 9 where x1 1 comma y 1 is 1 comma minus 9 so plug here then you will get minus 9 x minus 1 is equal to 3x square minus 9 now plugging the value of y the value of y is xq minus 9 so we'll write here x q minus 9 x q minus minus plus here plus 9 divided by x minus 1 is equal to 3x square minus 9 now multiply both side by x minus 1 by x minus 1 then you will get here x q minus 9x plus 9 is equal to x minus 1 into 3x squared minus 9 so we'll write here 3x squared minus 9 x q minus 9 x q minus 9 x q minus 9 plus 9 now multiply this one then you will get here 3x q minus 9x minus 3x squared 2x squared plus 9 now take all term right hand side here then you'll get here 3x q minus 9x minus 3x squared plus 9 now this is the xq in the plus so you'll write here minus x q plus 9x minus x q plus 9x now cancel this one also cancel 9x from here now we'll get here 3xqq x q is 2 x q minus 3 x squared now we can write like that 2 x q minus 3 x squared is equal to 0 now take the common x squared from the these these two terms then you will get x squared is equal to 2x minus 3 is equal to 0 now we have a two factors so either x square is equal to 0 or 2x minus 3 is equal to 0 to 0 from here the value of x is 0 and the from here the value of x is 3 by 2 now we have a 2 values of x here so we will get a 2 slope of the given curve so the slope of the this curve is 3 x squared minus 9 here now plug in x is equal to 0 then we'll get the value of slope is 3 into 0 square minus 9 is equal to minus 9 at x is equal to 0 so we can write here at x is equal to 0 now slope is minus 9 then we can compute the equation of the tangent line as you know the formula for the equation of the tangent line y minus y 1 is equal to m into x minus x1 where x1 y1 is 1 comma minus 9 so the value of y 1 is minus 9 here then we'll get y plus 9 the value of m is minus 9 x minus the value of x1 is 1 now we'll simplify this one then we'll get y plus 9 is equal to minus 9x plus 9 now subtract both sides 9 then we'll get to y is equal to minus 9 x this is the first equation of a tangent line now we'll compute the equation of a tangent line at x is equal to 3 by 2 at x equal to 3 by 2, the slope is 3 into 3 by 2 together whole square minus 9...

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