# Thread: Find the equation of the tangent line to y=1/2x^2 - 3 that has a slope of 4.?

1. ## Find the equation of the tangent line to y=1/2x^2 - 3 that has a slope of 4.?

So the two equations are
1) Y = x^2 + 1
2) Y = mx – 4
Equate both the equations
x^2+1 = mx -4
x^2 +5 + mx = 0
My teacher told me that from here use the quadratic formula in questions similar to these but in those questions the slope was given and the y-intercept wasn’t. so it worked out.. How do I find the value of m in the equations…

2. Originally Posted by ruscutie100
So the two equations are
1) Y = x^2 + 1
2) Y = mx – 4
Equate both the equations
x^2+1 = mx -4
x^2 +5 + mx = 0
My teacher told me that from here use the quadratic formula in questions similar to these but in those questions the slope was given and the y-intercept wasn’t. so it worked out.. How do I find the value of m in the equations…
Please ignore the topic..i edited it l8r on.. the question is
Find the equations of the tangent lines to Y = X^2 +1 that have a y intercept of -4

3. Originally Posted by ruscutie100
Draw the parabola.
Find the point of intersection.

So i was trying to do this question. My teacher told me that whenever you see f(x) = 1/x then assume that it's the reciprocal function and the graph always takes the same shape...here'z the link for the shape:-http://www.gomath.com/htdocs/images/7-10...

so shouldn't the equation y=1/2x^2 - 3 take the same shape... isn't it the reciprocal function because when i graph it on a graphing calculator it takes the shape of a parabola and i checked the answer the teacher gave us, and he also made it in a shape of parabola... can some1 explain me how do i come to know when is it reciprocal function and when is it a parabola
Plz help me asap.. i have a test on this tomorrow... thnx

(i don't know calculus)
Having a hard time figuring out your formula
y=1/2x^2 - 3

does this equal
a) $y= \frac{1}{2}x^{2}-3$

b) $y= \frac{1}{2x^{2}}-3$

c) $y= \frac{1}{2x^{2}-3}$

4. Originally Posted by angel.white
Having a hard time figuring out your formula
y=1/2x^2 - 3

does this equal
a) $y= \frac{1}{2}x^{2}-3$

b) $y= \frac{1}{2x^{2}}-3$

c) $y= \frac{1}{2x^{2}-3}$
srry i changed the questions... now i got it.. plz answer the edited one.. .thnx 4 replying 4 this question

5. $Y = X^{2} +1$ that have a y intercept of -4

Okay, so a y-intercept is a point where the x value is zero, thus intersecting the y-axis. So this means that our tangent lines have the point (0, -4) on them.

First, lets find the equation for our tangent lines. We know (or should) that $y\prime$ is our slope, this means that in the equation y=mx+b, m = $y\prime$

So first lets find $y\prime$

I'm assuming you understand how to differentiate,
$Y = X^{2} +1$

Becomes
$Y\prime = 2X$
If that step confused you, say so. You might also see this written as
$\frac{dy}{dx} = 2x$
They are the same thing.

So our tangent line has an equation of y=mx+b, and we now know that our slope = 2x for some x. We need to find that x in order to know our slope. We also know that to calculate a slope, we can take the difference in y of two points and divide by the difference in x. Well, we have one point of (0, -4), and we can calculate our other point as ( $x, x^{2}+1$) because we know that there is an x, y point on the parabola that is the same as on the line.

So
$\frac{y_1 - y_2}{x_1 - x_2} = slope$

replace the left side with our points, and the right side with 2x, because we know that 2x is our slope.
$\frac{(x^{2}+1) - (-4)}{x - 0} = 2x$

simplify
$\frac{x^{2} + 5}{x} = 2x$

multiply by x
$x^{2} + 5 = 2x^{2}$

subtract $x^{2}$
$5 = x^{2}$

square root
$\frac{+}{}\sqrt{5} = x$

plug that in for our slope of 2x to get slope = $\frac{+}{}2\sqrt{5}$

And now, FINALLY (lol) we can plug that into y=mx+b
so $y = \frac{+}{}2\sqrt{5}x+b$

And plug in our x, y values
$-4 = \frac{+}{}2\sqrt{5}(0)+b$
$-4 = b$

And substitute b in to our equation
$y = \frac{+}{}2\sqrt{5}x-4$

6. Originally Posted by ruscutie100
Find the equations of the tangent lines to Y = X^2 +1 that have a y intercept of -4
Hello,

the y-intercept is a point on the y-axis that means the x-value is zero.

1. Let this point be P(0,-4)
2. Calculate the first derivative which is the slope of the tangent:

$y=x^2+1~\implies~y'=2x$

3. The tangent point $T(t, t^2+1)$ is placed on the graph of the function too.

4. Plug in the coordinates of T:

$y'=2t = m$

Use the point-slope-formula of the equation of a straight line:

$2t=\frac{t^2+1-(-4)}{t-0}$ Solve for t. I've got: $t=-\sqrt{5}~\vee~t=\sqrt{5}$

5. Plug in these values:

a) The tangent points are $T_1(-\sqrt{5}, 6)$ or $T_2(\sqrt{5}, 6)$

b) the 2 tangents have the equations:
$\boxed{t_1:y=-2\sqrt{5} \cdot x-4}$ or $\boxed{t_2:y=2\sqrt{5} \cdot x-4}$

7. Okay, sorry that took so long, I had to get some things straight in my head.

It should be correct now, I won't edit it again.