tangents to y=4x^5-3x^4+15x^2+6 passing thru origin

[xc630]

At how many points on the curve y=4x^5-3x^4+15x^2+6 will the line tangent to the curve pass through the origin?

I am not totally sure what to do. On one one hand I am thinking the answer is one because how could a single line pass through a point more than once. However if thee is more than one point I am not sure how to find the tangent line if a point on the curve is not given. Soem help please!

 

[stapel]

The exercise doesn't ask you for the number of times that one particular line might pass through the origin. It asks you for the number of tangent lines to the given curve will pass through the origin.

Yes, each such tangent line will cross the origin only once. But that says nothing about how many of those tangent lines there might be.

First, you've done a graph (at least in your calculator), right? What do you see from that?

Thank you.

Eliz.

 

[xc630]

in the third quadrant the curve is concave down and smooths out at the x-axis as you go right. it then appears to pass through the origin and then forms a concave up curve in the first quadrant. From that I think only one line that is tangent passes thorugh the origin

 

[galactus]

Start by finding the derivative(m=slope) of your polynomial. Do as Stapel said and graph the function. It will help you get an idea where they are. Close.

Since the tangent lines pass through (0,0), they will have equation y=mx.

You are not asked to find the actual equations, though.

You can use \(\displaystyle \L\\y-y_{1}=m(x-x_{1})\)

You're given y_1=0 and x_1=0.

\(\displaystyle \L\\\underbrace{4x^{5}-3x^{4}+15x^{2}+6}_{\text{y}}-\underbrace{0}_{\text{y_{1}}}=\underbrace{m}_{\text{f'(x)}}(x-\underbrace{0}_{\text{x_{1}}})\)

The real roots of the resulting function will be good ones.

You can use the Intermediate Value Theorem to hone in on it/them.

I used technology to solve. Do you have a nice calculator?.