Argument of a function - what? (1 Viewer)

[Symphonicity]

Hello,

I'm looking at Mendelson's "Beginning Calculus" and having real trouble making sense of this question.

"For each function f and argument x = a below, find the slope-intercept equation of the tangent line corresponding to the given argument a."

The function is : f(x) = 2x^2 + x a = 1/4.

The answer is y = 2x - 1/8. Please could someone explain about the "argument a"? I know that an argument is meant to be an input into a function but having trouble working out how to come to that answer. Thanks very much.

 

[pdang]

you're basically just finding the equation of the tangent at x=1/4

 

[Symphonicity]

i eventually got it. thanks. Can I ask a supplementary question from the same set - "find the point(s) on the curve y=x^2 at which the tangent line is passing through the point (2, -12)"

So since (2,-12) is not on the curve, I guess this is asking me to work out where the tangent line comes into contact with the curve, knowing that it passes through that point. Any hints??

Thank you.

Edit: I've tried this method and got an answer. Does this work:

We know that the slope of a tangent at a given point on y=x^2 is 2x. So the tangent has the equation of y=mx+b and m must equal 2x.

So I can put that in here:

y-y1 = m(x-x1)

y+12 = 2x(x-2) (I've subbed in the x and y values for the given point in the question).

Then I can substitute y = x^2 (but I don't really understand why this works)

and I get x^2 + 12 = 2x^2 -4x.

Simplifies to x^2 - 4x - 12 = 0.

becomes (x-6)(x+2) = 0.

I get values of 6, -2 for x. Which gives points (6,36) and (2,-12) that the line passes through. This answer is correct, my only quarrel is why can I substitute y=x^2 into the equation of the tangent line?

Thanks heaps.

 

[pdang]

the tangent must pass through 2 points: (2, -12) and (a, a^2). it must also have a gradient of 2a.
try putting these two conditions together

 

[cookiez69]

i eventually got it. thanks. Can I ask a supplementary question from the same set - "find the point(s) on the curve y=x^2 at which the tangent line is passing through the point (2, -12)"

So since (2,-12) is not on the curve, I guess this is asking me to work out where the tangent line comes into contact with the curve, knowing that it passes through that point. Any hints??

Thank you.

Edit: I've tried this method and got an answer. Does this work:

We know that the slope of a tangent at a given point on y=x^2 is 2x. So the tangent has the equation of y=mx+b and m must equal 2x.

So I can put that in here:

y-y1 = m(x-x1)

y+12 = 2x(x-2) (I've subbed in the x and y values for the given point in the question).

Then I can substitute y = x^2 (but I don't really understand why this works)

and I get x^2 + 12 = 2x^2 -4x.

Simplifies to x^2 - 4x - 12 = 0.

becomes (x-6)(x+2) = 0.

I get values of 6, -2 for x. Which gives points (6,36) and (2,-12) that the line passes through. This answer is correct, my only quarrel is why can I substitute y=x^2 into the equation of the tangent line?

Thanks heaps.

It's because you're finding the point of which the tangent touches the curve. By subbing in y = x^2 into the equation of the tangent, you find out the x-values of where the line is a tangent to the curve. It's just like how you use simultaneous equations to figure out where two graphs intersect one another.

 

[Symphonicity]

It's because you're finding the point of which the tangent touches the curve. By subbing in y = x^2 into the equation of the tangent, you find out the x-values of where the line is a tangent to the curve. It's just like how you use simultaneous equations to figure out where two graphs intersect one another.

Ah perfect, thank you. I'm happy now. So why call the line a tangent if it intersects the curve in two places? Wouldn't that make it a secant?

 

[cookiez69]

Ah perfect, thank you. I'm happy now. So why call the line a tangent if it intersects the curve in two places? Wouldn't that make it a secant?

No. There are actually two lines that pass through the point and are a tangent to the curve at points (6,36) and (2,-12). If you figure out the equation of the two lines, you'll see that one has a positive gradient and the other has a negative gradient.