# Homework Help: Find the equations of 2 tangents to a curve with a POI

1. Oct 27, 2016

### Arnoldjavs3

1. The problem statement, all variables and given/known data
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2. Relevant equations

3. The attempt at a solution
Okay, so the things I know before hand is that f'(x) = 4x^3
however, my question is since the problem states that the line that passes through (-1.25, -8) is tangent to the curve, wouldn't it have to satisfy f(x) = (-1.25)^4 = -8?
If it doesn't, that means it's not actually tangent to the curve? (I believe this is where my misunderstanding is founded from)

How can I visualize this question? I'm having a hard time answering it without having a grasp of what it could look like. The hint just confused me even more

Thanks

2. Oct 27, 2016

### PeroK

You need to draw a graph of the function $f(x) = x^4$ and the point $(-1.25, -8)$.

You could get an approx answer from this graphically.

Then, you can apply the algebra to get a precise solution.

Note that a line tangent to a curve is not the curve! It only touches the curve at a single point.

3. Oct 28, 2016

### Arnoldjavs3

I'm aware of what a tangent is, but(please don't take this the wrong way) am I not to use any calculus here?

If it matters, the homework problems I was provided with is about the intermediate value thereom. I'm having a hard time relating that to what's going on here.

Last edited: Oct 28, 2016
4. Oct 28, 2016

### PeroK

You don't need calculus - not more than the derivative you alraedy have. Have you drawn a diagram? What distingishes a tangent from the other two options for a line relative to the curve?

PS the hint tells you what to do. How far can you get with this problem?

Last edited: Oct 28, 2016
5. Oct 28, 2016

### Arnoldjavs3

Using the hint as you directed,

I can single out $$b=4a^4 + 5a^3 + 8$$

Since the equation of the tangent lines is $y = m(x-x1) + y$, I would assume that I would substitute the slope $4a^3$ for m and then use a and b as points?

$$y=4a^3(x-a)+(4a^4 + 5a^3 - 8)$$
Am I on the right track here? If so,

$$4a^3x + 5a^3 - 8$$ is what I end up with, however I don't know what to do past this point(i'm assuming it's false to begin with)

6. Oct 29, 2016

### PeroK

This is what you were given by the hint. All you needed to do was notice that $b = a^4$ (as this is a point on the curve $y = x^4$) and then solve the resulting quartic equation.

So, let me try to explain.

1) You have a specific point $(-\frac{5}{4}, -8)$.

2) You want to consider all lines through this point. Let the slope of a line be $m$. The equation of this line is then:

$y + 8 = m(x + \frac{5}{4})$

3) Some of these lines will not intersect the curve at all, and some will intersect the curve, either cutting through the curve or, in two cases, just touching the curve. If you drew a diagram, you would see this immediately and also see rougly what the slope $m$ must be for the line to be a tangent.

4) Where this line interects the curve you also have $y = x^4$. So, for all points of intersection $(a, b)$ you have:

$b + 8 = m(a + \frac{5}{4})$ and $b = a^4$

Hence:

$a^4 + 8 = m(a + \frac{5}{4})$

5) In addition, if the point of intersection is a tangent point, then the slope of the line $m$ equals the slope of the curve $4a^3$ at that point. This gives an equation for any tangent points:

$a^4 + 8 = 4a^3(a + \frac{5}{4})$

Which simplifies to:

$3a^4 + 5a^3 - 8 = 0$

Now, all you have to do is solve this equation.

Does this make any sense?

7. Oct 29, 2016

### Arnoldjavs3

I actually did end up solving this problem, but it is useful to know how the line with point(-1.25, -8) may or may not cross the curve(in this case it doesn't). Yeah I do understand what you're saying, it's just that on the midterm that I have upcoming I won't be allowed a calculator / paper to graph so I wanted to try solving it using only algebra.

The equations were $$y=-32(x+1)$$ and $$y=4x-3$$ from what I had gotten.

Err, is there a specific reason why I can't edit my posts from before?