Archive for February, 2013

Question:

If f(x)=x^2-10, find an equation of the tangent to the curve passing through the point (5, 1).

Answer:

f'(x)=2x

f'(5)=10

y=10x+b

1=10(5)+b

b=-49

\therefore y=10x-49

Congratulation if you put this answer on your test or exam!  Zero mark!

Because it is wrong, MEOW…

wrong.answerUsually students will say,

“What’s wrong?  This is how my teacher taught us in class.”

“I think the answer is close enough.”

“I don’t care I think I can get part mark for my work.”

“Show me where my mistake is, I think my answer is perfectly right.”

“My textbook did it this way.”

…, usual BS, etc….

Those are just some responses from students I have encountered in the past.  It isn’t hard to quickly show the above answer is wrong.  One may simply do this,

x^2-10=10x-49

x^2-10x+39=0

then the determinant of the quadratic equation is,

10^2-4(1)(39)=100-156=-56

No real solution, so the straight line and the parabola don’t intersect at all, no way it is a tangent.

The problem of the above solution is that the student assumed the point (5, 1) is on the curve which it isn’t.

So the correct answer should be the following.

Correct Answer:

Obviously the point (5, 1) is not on the curve since,

5^2-10=15\neq1

hence the point (5, 1) only lies on the straight line that is tangent to the parabola.

Let the tangent touches the parabola at a point (x, y), and since f'(x)=2x.  By the point slope form of a straight line,

y-y_0=m(x-x_0)

We have,

y-1=2x(x-5)

y=2x^2-10x+1

Now since the tangent touches the parabola at one point,

2x^2-10x+1=x^2-10

x^2-10x+11=0

x=\dfrac{10\pm\sqrt{100-44}}{2}=\dfrac{10\pm\sqrt{56}}{2}=5\pm\sqrt{14}

Therefore the slope of the tangent are 2x=10\pm2\sqrt{14}, there are two possible tangent lines.

They are

y-1=(10+2\sqrt{14})(x-5)

and

y-1=(10-2\sqrt{14})(x-5)

Done.

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