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…
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 is our slope, this means that in the equation y=mx+b, m =
So first lets find
I'm assuming you understand how to differentiate,
If that step confused you, say so. You might also see this written as
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 ( ) because we know that there is an x, y point on the parabola that is the same as on the line.
replace the left side with our points, and the right side with 2x, because we know that 2x is our slope.
multiply by x
plug that in for our slope of 2x to get slope =
And now, FINALLY (lol) we can plug that into y=mx+b
And plug in our x, y values
And substitute b in to our equation
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:
3. The tangent point is placed on the graph of the function too.
4. Plug in the coordinates of T:
Use the point-slope-formula of the equation of a straight line:
Solve for t. I've got:
5. Plug in these values:
a) The tangent points are or
b) the 2 tangents have the equations: