Let ( f(x)=x^3-2 x^2+a x+6 ). Part of the graph of ( f ) is...

Let ( f(x)=x^{3}-2 x^{2}+a x+6 ). Part of the graph of ( f ) is shown in the following diagram. The graph of ( f ) crosses the ( y )-axis at the point ( mathrm{P} ). The line ( L ) is tangent to the graph of ( f ) at ( mathrm{P} ). (a) Find the coordinates of ( mathrm{P} ). ( [2] ) (b) (i) Find ( f^{prime}(x) ). (ii) Hence, find the equation of ( L ) in terms of ( a ). [6] The graph of ( f ) has a local minimum at the point ( mathrm{Q} ). The line ( L ) passes through ( mathrm{Q} ). (c) Find the value of ( a ). [8]

Transcript

00:01 In this question, we are given the function f, and this is the curve, and this is the function, and we are given that l is tangent to the function f at point p.

00:18 In part a, we want to find the coordinates of p.

00:21 So at point p, x is 0, so just sub in 0 into the function, function.

00:31 So f0 will be 0 minus 0 plus 0 plus 6.

00:37 So the y value or y in the set for function f is 6.

00:42 So therefore p is 0, 6.

00:46 That's the coordinate.

00:47 So this point here is 0, 6.

00:56 B part 1, we want to find f prime x.

00:59 That is differentiate f respect to x.

01:02 Now since there are four terms here separated by minus and plus sign, we can do term by term differentiation.

01:09 For the first term x cubed, when we differentiate that, we'll get bring down the power 3, repeat the x and subtract 1 to the power.

01:18 Now 2 is multiplied variable, so it's kept aside, differentiating x square, bring down the power 2, repeat the x and subtract 1 to the power.

01:27 Now a is a constant, multiplied to x, so it's kept cap aside, an x -differentiated aspect to itself is just 1.

01:34 And 6 is a constant, when differentiated, you get 0.

01:38 So f prime x is 3x squared minus 4x plus a.

01:47 Part 2, we want to find the equation of l at the point p.

01:53 Now, since l is a tangent at the point p, the gradient of line l is equal to the differentiation of the function f at the point p.

02:16 And that is when x is 0.

02:21 So f prime 0 is just sub 0 into the x, i'll get 0 minus 0 plus a.

02:28 So the gradient of the line is a.

02:31 Now in general, for a line with gradient m and passes through a point x0, y0, the equation of the line would be y minus y0 equals to m bracket x minus x0.

02:49 So the equation of line l at point point p where p is 0, 6...

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