Yr12 Math Methods | Need Help Understanding The Way This Has Been Solved.

[Brittank88]

Attached is official solution to the question I'm having trouble with.

Putting aside the fact I have no idea how to solve it myself, I can't even understand what they've done to solve it either.

Hoping someone can explain to my the how and the why.

[Lear]

They’ve basically used the linear equation form y-y1=m(x-x1) to find the tangent. y1 is on the first line by finding f(a). m is found by differentiating f(a). And we know from the question that x1 is a.
 On line 3 the y-y1= m(x-x1) step can be clearly observed.

Hope it helps

[jazzycab]

Attached is official solution to the question I'm having trouble with.

Putting aside the fact I have no idea how to solve it myself, I can't even understand what they've done to solve it either.

Hoping someone can explain to my the how and the why.

Firstly, the question must obviously introduce the function \(f\left(x\right)=-10x^2\left(2x-3\right)^3\) in some earlier part.
To find the equation of the tangent at any point, we need to know the gradient of the curve at the point, hence, we differentiate \(f\left(x\right)\) using the product rule (note that they substituted in \(x=a\) first, which is not advisable, as this technique won't work when substituting an actual number in).
The value obtained for \(f'\left(a\right)\) is therefore the gradient of the tangent line at \(x=a\), i.e. \(m=-20a\left(2a-3\right)^2\left(5a-3\right)\).
To find the equation of any line, we need it's gradient and a point on the line. The \(x\)-value of the point in question is \(a\), the \(y\)-value can be obtained by substituting \(x=a\) into \(y=f\left(x\right)\), giving \(y=-10a^2\left(2a-3\right)^3\).
The equation of a line can be found by substituting the point \(\left(x_1, y_1\right)\) and the gradient, \(m\) into the equation \(y-y_1=m\left(x-x_1\right)\), which gives the equation given.

[Brittank88]

Thanks guys (& girls), makes much more sense now.

My main issue was that I only knew the form , and I don't remember ever covering the other linear form you all mentioned (which will be very helpful in the future, so thank you).

[jazzycab]

Thanks guys (& girls), makes much more sense now.

My main issue was that I only knew the form , and I don't remember ever covering the other linear form you all mentioned (which will be very helpful in the future, so thank you).

Note that you will get the same answer even if you use that form of a straight line equation