Question

Find an equation of the normal line to the parabola y = x2 − 8x + 4 that is parallel to the line x − 4y = 5.

y =

Answer #1

There is a line through the origin that divides the region
bounded by the parabola y=8x−7x2 and the x-axis into two
regions with equal area. What is the slope of that line?

Consider the parabola given by the equation
y = x2 −
5
4
. Let
(a,b)
be the point lying on this parabola, with
a ≥ 0
, which is closest to the origin.
What is
a2 / b2
?
(The solution, as usual, is a whole number.)

Find an equation for the line tangent to the parabola y= 2x^2
-13x +5 which has a slope of -1

Find the volume generated by revolving the area bounded by the
parabola y^2 = 8x and its latus rectum about the latus rectum. Use
the shell method.

Consider f(x) = x2 – 8x. Find its derivative using
the limit definition of the derivative. Simplify all
steps.
a. Find f(x + h).
____________
b. Find f(x + h) – f(x).
____________
c. Find [f(x + h) – f(x)] ÷ h.
____________
d. Find lim (hà0) [f(x + h) – f(x)] ÷ h.
____________
e. Find an equation of the line tangent to
the graph of y = x2 – 8x where x = -3. Present your
answer...

1)
Find the equation of the line that psses through the vertex if the
parabola f(x)=2x^2-12x+19 and that crosses the x-axis at x=5
2) Find the equation of the line that passes through the
certex of the parabala f(x)=3x^2-12x+17 and has y-intercept of
(0,10)

Find an equation of the line that is tangent to the graph of
f and parallel to the given line.
Function
Line
f(x) = x2
6x − y + 9 = 0

Find the equation of the osculating
circle of the parabola y =
x^2 at the
origin.

Find a parabola with equation y = ax2 + bx + c that has slope 1
at x = 1, slope −11 at x = −1, and passes through the point (1,
4).

Solve the differential equation by UC Method. Do not evaluate
the exact value of coefficients.
y^'''+y^''-4y^'-4y=8x+8+6e^-x

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