Question about Office Equipment & Supplies

# Quadratic functions whatis the equation of te quadratic function if the vertex is 8,-4 and a= -2

Posted by on

• Level 3:

An expert who has achieved level 3 by getting 1000 points

Superstar:

An expert that got 20 achievements.

All-Star:

An expert that got 10 achievements.

MVP:

An expert that got 5 achievements.

• Office Equip... Master

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

Posted on Nov 21, 2013

Hi,
A 6ya expert can help you resolve that issue over the phone in a minute or two.
Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
The service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
Good luck!

Posted on Jan 02, 2017

(a-b)3

Posted on Apr 04, 2009

×

my-video-file.mp4

×

## Related Questions:

### 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

### I want to solve an complex equation on casio fx-570 ES Plus

Your scientific calculator is unable to solve complex equation with complex coefficients. You should try to solve by hand directly using the quadratic formula or by factoring the polynomial in z
failing that, another way would be to set z=x+iy, substitute this for z, carry out the algebra and try to separate real and imaginary parts. But your two equations will constitute a system of two quadratic equations. I am not aware of any general method to solve coupled nonlinear equations.
Good luck.

Dec 20, 2012 | Casio FX-115ES Scientific Calculator

### How do i use this for algebra equations, putting in varables and such

If you are just trying to solve a simple quadratic equation by entering coefficients, this calculator does have those features. The TI-83 is a better choice for that kind of function.

Sep 15, 2011 | Texas Instruments TI-30 XIIS Calculator

### How can I compute quadratic equations using fx-300ES?

You cannot solve the quadratic equation with this calculator. It does not have a solve program.

However if you wnat to create a table of values for a function (any type of function that uses the basic scientific function keys ) you use the TABLE Computational Mode.

You press MODE
You select TABLE
You enter a mathematical expression 3X^2-.33X+ 15
You generate the table of values.

Aug 28, 2011 | Casio fx-300ES Calculator

### Is there an example for experiments involving quadratic functions?

I am not sure what you mean by experiment. If you are wondering if the quadratic equation is useful or describes some physical process, the answer is yes: The simplest case is the description of the trajectory of a projectile in vacuum.
If you can be more specific I will try to help.

Aug 17, 2011 | As Seen On TV Office Equipment & Supplies

### I am trying to solve a quadratic function using the calculator but i dont seem to be able to do so. For example if i had a question like 5x^2 + 15x +3, how would i use the sharp el-w535 to find the x...

This expression 5x^2+15x+3 is not an equation, therefore it can have any value depending on the value of x. You have to make it an equation before looking for the particular value of x that satisfy your equation 5x^2+15x+3 =0

This calculator cannot solve equations because it does not have an equation solver.

Dec 15, 2010 | Sharp ELW535 Calculator

### Functions?

Hello,
I am afraid I am missing something here. You can start with the expression of a function and have it graphed on a graphing calculator.
Or you can enter the experimental data, represent them on a graphing calculator and ask the calculator to find the curve that fits the data best, and once that curve is found to give you the equation.
However your calculator does not have the graphing capabilities. All you can do is enter the data and ask the calculator to perform a regression to obtain the expression of the function that fits best.

The available options for the calculator are accessible by pressing [MODE][3:Stat] and
2. A+BX linear
4: ln X logarithmic
5: e^X exponential
6:A*B^X power
7: A*X^B Polynomial
8: 1/X

Hope it helps.

Apr 14, 2009 | Casio fx-300ES Calculator

### 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

## Open Questions:

#### Related Topics:

109 people viewed this question

Level 3 Expert

Level 3 Expert