Category Archives: MAC1140

Example: Find the equation of the parabola described, find the two points that define the latus rectum, and graph the parabola.

Focus at (0,-3) and vertex at (2,-3)

Solution:

I would like to figure out the orientation of the parabola so that I can decide which format to use when building the equation. I will put the information that I was given on a graph to help me figure out the orientation.

The focus is always inside the parabola and the vertex is a turning point. This tells me that the parabola must open to the left.

Since the parabola opens to the left, the equation of the parabola is as follows.

(y-k)^2=-4a(x-h)

See the post about the different equations of a parabola. http://math15fun.com/2017/07/05/features-of-a-parabola/

To write the equation, I must know the vertex: (h,k) and a which is the distance between the vertex and the focus.

The vertex is given in the instructions (h,k)=(2,-3) and I can find the distance between the vertex and focus by counting the units between. Thus a=2 .

I will fill these values into the formula and simplify.

(y-k)^2=-4a(x-h)

(y-(-3))^2=-4(2)(x-2)

(y+3)^2=-8(x-2)

The equation of the parabola with focus at (0,-3) and vertex at (2,-3) is

(y+3)^2=-8(x-2) .

The latus rectum points are points on the parabola that are in line with the focus.

By my drawing, you notice that the two points that define the latus rectum and the focus share the same x-coordinate. This is only true when the parabola is oriented horizontally as in this example. The latus rectum points will share the y-coordinate with the focus when the parabola is vertically oriented.

The x-coordinate of the latus rectum points will be 0 and since the latus rectum points are on the parabola, I can use the equation of the parabola to help me find the rest of the ordered pairs.

Substitute a 0 in for x in the equation of the parabola and solve for y.

The equation of the parabola.
Substitute 0 in for x and simplify.
Solve for y by getting rid of the square by taking the square root both sides and simplifying.	$sqrt{(y+3)^2}=sqrt{16}$
Solve for y by getting rid of the plus 3 on both sides by subtracting 3 on both sides and simplifying.
Write the plus or minus symbol separately and simplify.	and and

The latus rectum is defined by the points. (0,1) and (0,-7) .

To draw the graph, place the vertex and the latus rectum points on the graph.

The latus rectum points show you how wide to draw the parabola.

7.2 The Parabola

Features of a Parabola

July 5, 2017 admin

A parabola is the set of all points P in a plane that have the same distance from a fixed point F (the focus) as they are from a fixes line D (the directrix).

Orientation: There are four main orientations of the parabola which are up, down, right and left.

Equation:

Orientation	Graph	Verbal Description of the Equation	Equation
Up		Vertical orientation will have the square on the x variable.
Down		Vertical orientation will have the square on the x variable and will have a negative because of the reflection.
Right		Horizontal orientation will have the square on the y variable.
Left		Horizontal orientation will have the square on the y variable and will have a negative because of the reflection.

Features of A Parabola Oriented Up

Study the vocabulary and the location of these important features of a parabola.

3.1 Functions, 3.1 Relations and Functions, Difference Quotient

Difference Quotient: Quadratic Function

June 26, 2017 admin

Example: Find the difference quotient for f(x)=x^2-9x

The Difference Quotient: {f(x+h)-f(x)}/h

Solution:

The Difference Quotient Formula
Write the difference quotient for the given function	= ${(x+h)^2-9(x+h)-(x^2-9x)}/h$
Apply the exponent and use the distributive property	= ${(x+h)(x+h)-9x-9h-x^2+9x}/h$
Multiply	= ${x^2+xh+xh+h^2-9x-9h-x^2+9x}/h$
Combine the like terms	= ${x^2+2xh+h^2-9x-9h-x^2+9x}/h$
Combine the like terms. Only terms with h should remain	= ${2xh+h^2-9h}/h$
Divide h into each term	= ${2xh}/h+{h^2}/h-{9h}/h$
Cancel the common h from each term	=