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Determining the equation of the parabola

Find the equation of the parabola with the vertex of (2,0) and focus of (2,2)

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V (2,0) and Focus (2,2)
since the focus is 2 units above the vertex, the parabola opens upward

vertex (h,k)
a is the focal length (distance between the vertex and the focus)
a = 2
(2 units above the vertex)

(x-h)^2 = 4a (y-k)
(x-2)^2 = 4(2) (y-0)
x^2 - 4x + 4 = 8 (y)
x^2 - 4x + 4 = 8y
x^2 - 4x - 8y + 4 = 0

**Answer: "(x-2)^2 = 8 (y)" or in expanded form "x^2 - 4x - 8y +4"

Posted on Jan 10, 2009

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Whats the answer to y=(x+2)^2 using a new equation in vertical form a transfer right 4 unit up 2 unit and reflect on x axis


Let's start with the original equation. It is in the form y=a(x-h)^2+k, where (h,k) is the vertex. In this case, h= - 2 and k= 0. Thus, is it y=x^2 and we have done a horizontal translation 2 units to the left. I assume we are going to translate this 4 units to the right, then 2 units up, and then reflect it in the x-axis. Before we do the reflection, we will be 2 units to the right and 2 units up. This equation would be y=(x-2)^2+2. If we reflect this in the x-axis, would we get a downward facing parabola y=-(x-2)^2 -2?

Good luck,

Paul

Dec 06, 2015 | Office Equipment & Supplies

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How i can solve quadratic equations which have no real roots


If the quadratic equation has no roots, you cannot find the roots. The discriminate is negative, so if we attempt to use the quadratic equation, we get no roots.

For example, y=x^2 +0x + 3

a=1, b=0, c=3

(-b+/- sqrt(b^2 -4ac))/2a

Substituting in the numbers, we get

x= (-0 +/-sqrt(0^2 - 4(1)3))/2
x= (0 +/- sqrt (-12))/2

We cannot do the square root of -12. Therefore, there are no roots. This is the same as having no x-intercepts. The discriminant is b^2-4ac. In this case it is -12. Thus, there are no real roots.

However, you can still determine the maximum or minimum, the vertex, the axis of symmetry, the y-intercept and the stretch/compression. With this information you can graph the equation.

In this case, the y-intercept is 3, the vertex is at (0,3), the axis of symmetry is x=0, the minimum value of the function is 3.

Good luck.

Let me know if you have any questions.

Paul

Nov 04, 2015 | Casio FX991ES Scientific Calculator

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Determine the domain and the range of the


Domain and range are concepts that many students have trouble with, so you are not alone!

Domain is the x values of the relation.
Range is the y values of the relation.

If you have trouble remembering which one is which , like I do, D comes before R, just like x comes before y;)

To determine if a relation is a function, there are several tests that you can use. For every x value, there can be one and only one y value. If there is more than one y value for any x value, it is not a function.

Another test is the vertical line test. If a vertical line only goes through one point on the line, it is a function. If it goes through two points on the line, it is not a function.

As an example, let's look at y = (x-3)^2 + 6.

This parabola is in the form, y=a(x-h)^2 + k, where a indicates a stretch or compression, and whether the parabola opens up or down. If a is positive, it opens up, and the y value of the vertex represents a minimum. If a is negative, it opens down, and the y value of the vertex is a maximum. The values h and k are the x and y values of the vertex.

In this case, a is one, so the parabola opens up, with a minimum value of 6.''

Now back to the domain and range.

Since we can use any value for x in the equation, for the domain, x is an element of real numbers, sometimes written (xER).

For the range, y can only be greater than or equal to the minimum. So the range is y is an element of all real numbers, such that y is greater or equal to 6, sometimes written (yER'y>=6). Sorry I couldn't find the greater than or equal to character;)

Good luck.

Paul

Oct 03, 2014 | Office Equipment & Supplies

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What is the line of symmetry for the parabola whose equation is y = x2 + 10x + 25?


Factor the expression,
x^2+10x+25=(x+5)^2
The minimum is the point ( x=-5, y=0)
The axis of symmetry is the vertical line with equation x=-5

May 08, 2014 | Office Equipment & Supplies

1 Answer

What is the axis of symmetry of the function f(x) = x2 + 10x?


It is the vertical line that passes through the vertex of the parabola y=x^2+10x
You can graph the function or use differential calculus to find position of the minimum.
The vertex is at (-5,-25) and the axis of symmetry is the line x=-5.

Dec 19, 2013 | Office Equipment & Supplies

1 Answer

Quadratic functions


Y=aX^2+bX+c

The x-coordinate of the vertex is given by
X_v=-b/2a
With a=-2, and X_v=8 you have b=-2a*X_v=-2(-2)*8=32
The equation can now be written as y=-2X^2+32X+c
To get c you substitute the value of X_v in the equation above and you set the result equal to Y_v=-4
-2(8)^2+32(8)+c=-4
Solve for c
c=-4+128-256=-132
y=-2X^2+32X-132

Nov 21, 2013 | Office Equipment & Supplies

1 Answer

My fx-9750GA PLUS wont graph lines, cicles or paraballas


Sure it does! Did you ask it to graph?Here are some screen captures to help you find your way.
To draw functions (regular, parametric, polar, sequences) use the application graph: Highloghted on 1st screen capture. The second screen captures shows you the functions y=f(x). enter a legal expression and press F6 Draw. To set TYPE press the key under TYPE.

k24674_100.jpgk24674_101.jpg

To draw conics (circles, parabolas, hyperbolas, and ellipses) use the CONiCS applications. (See 3rd screen ). On 4th capture the equation highlight is one for the parabola. y=x^2

k24674_102.jpgk24674_103.jpg

Scroll down or up examining the small graphs. They will show you which equation form to choose.
For the circles
k24674_104.jpg

Jun 19, 2011 | Casio FX-9750GPlus Calculator

1 Answer

How do i find where the x value will be 15 on a parabola


If you have a function for the parabola it is in the form y = a x^2 + b x +c. To find what y is when x = 15 plug 15 in every time there is an x. so the equation is in the form y = a (15)^2 + b (15) + (15) .

Mar 08, 2011 | Texas Instruments Office Equipment &...

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TI-83 graph problems-help please. Hi, sorry to sound so stupid, but trying for hours to try to solve these using my calculator,but doing something wrong as can't get the right answers. I wonder if anyone...


Parabolas
Open Y= editor and type in the two functions
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The calculaus functions are accessible by pressing [2nd][TRACE] to open the CALCulate menu options. For the gradient (I think you mean the derivative) use option 6:dy/dx. But first choose the point where you want it calculated (use cursor to move along the curve) and press ENTER. The value of the deivative will be calculated at the chosen point.
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The vertex of the parabolas are maxima. Thus you must use option 4:maximum
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You will be prompted for a left bound. Move cursor to the left of the maximum (not too far) and press [ENTER]. A fat arrow is displayed on the graph that shows the left limit of the interval. You will be prompted for a right bound. Move cursor along the curve to the right of the the vertex. Press ENTER. A seconf fat arrow will be displayed to show the right limit of the interval.

You will be prompted for a guess of the maximum. Move cursor newar the max or enter a numerical value and press ENTER.

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The location of the vertex is displayed (X and Y values)
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I have no idea what you mean by the equation of symmetry

Intercept.

May 02, 2010 | Texas Instruments TI-83 Plus Calculator

1 Answer

How do I graph parabolas?


Parabolas are forms of the base equation X^2. So press the [Y=] key, and then type in the variation of x squared that you want

May 11, 2009 | Texas Instruments TI-83 Plus Calculator

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