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

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X2 + 10 x + 25 = 2 (x+5)2 = 2 x+5 = +/- sqrt(2) x = -5 +/- sqrt(2)

Posted on May 08, 2014

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SOURCE: when i put an equation

The truth of the matter is that the curve drawn (the parabola) is the graph that you are seeking.
It is possible that you you mean by graph is the the set of axes. In that case you may have deactivated them. To reactivate them:

  1. Press [2nd][ZOOM] to access the (FORMAT).
  2. Press the DownArrow key 3 times to reach the line AxesON AxesOFF.
  3. Use the rightArrow to highlight AxesON and press [ENTER].
  4. Your graph will have the axes displayed.

Posted on Aug 04, 2010

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1 Answer

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

Write an equation of the line in standard form that pass through (-5,-11) and 10,7)


Calcualte the slope of the line as
a=(7-(-11))/(10-(-5))=18/15=6/5
Use the fact that the line passes through one of the two points, for example (10,7)
7=(6/5)*10+b=12+b
Obtain b as b=7-12=-5
The equation of the line in functional form is y=(6/5)x-5
Multiply everything by 5 to clear the fraction
5y=6x-25 or 0=6x-5y-25
Finally, the equation in general form (standard?) is 6x-5y-25=0.

Check the calculation by verifying that the point (10,7) lies on the line.
6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!
Check that the second point (-5,-11) lies on the line also (if you want to)
6*(-5)-5*(-11)-25=-30+55-25=0
That checks OK.

Dec 04, 2011 | Super Tutor Pre Algebra (ESDPALG)

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 &...

1 Answer

When i put an equation in Y= the parabola appears but not the graph. why? it was just fine


The truth of the matter is that the curve drawn (the parabola) is the graph that you are seeking.
It is possible that you you mean by graph is the the set of axes. In that case you may have deactivated them. To reactivate them:
  1. Press [2nd][ZOOM] to access the (FORMAT).
  2. Press the DownArrow key 3 times to reach the line AxesON AxesOFF.
  3. Use the rightArrow to highlight AxesON and press [ENTER].
  4. Your graph will have the axes displayed.

Aug 04, 2010 | Texas Instruments TI-83 Plus Calculator

2 Answers

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
6447ca5.jpgbbeaa23.jpg
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.
50e3257.jpg005508b.jpg

The vertex of the parabolas are maxima. Thus you must use option 4:maximum
24d4168.jpgb57bf8d.jpg
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.

3b702ee.jpg22c11ef.jpg

The location of the vertex is displayed (X and Y values)
3908f7c.jpg
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

1 Answer

Determining the equation of the parabola


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"

Nov 18, 2008 | Super Tutor Trigonometry (ESDTRIG) for PC

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