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

See image below

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

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.

Calculator manual: http://education.ti.com/downloads/guidebooks/scientific/30xii/30xiiqrg-eng.pdf
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.

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.

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.

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.

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
3: _+cX^2 quadratic
4: ln X logarithmic
5: e^X exponential
6:A*B^X power
7: A*X^B Polynomial
8: 1/X

Hope it helps.

(a-b)3

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"