quadratic functions and parabolas

[lriosmo]

Help! I'm lost. The question is: a. find coordinates of the vertex for the parabola b. use vertex and intercepts to sketch a graph c. give the equation of the parabola's axis of symmetry and d. use the graph to determine the function's domain and range.
Here are two quadratic functions. (I include them both because they look different and I don't know what to do with each form):

1. f(x) = 2 (x+2)[sup:2luvss0y]2[/sup:2luvss0y]-1

2. f(x) = 2x[sup:2luvss0y]2[/sup:2luvss0y]+ 4x -3

Thank you!

 

[tutor_joel]

there are a number of ways of doing this. However, for your needs, this is done by understanding transformations.

fist of all if the equations in vertex form, which eq1 is in, then the equation for the vertex is simply x = h. The vertex form is y = a(x-h)^2 + k.

If the equation is in standard form ie. #2. ax^2 + bx + c, then the equation for the vertex is x = -b/2a

 

[masters]

Help! I'm lost. The question is: a. find coordinates of the vertex for the parabola b. use vertex and intercepts to sketch a graph c. give the equation of the parabola's axis of symmetry and d. use the graph to determine the function's domain and range.
Here are two quadratic functions. (I include them both because they look different and I don't know what to do with each form):

1. f(x) = 2 (x+2)[sup:3abhgxo7]2[/sup:3abhgxo7]-1

2. f(x) = 2x[sup:3abhgxo7]2[/sup:3abhgxo7]+ 4x -3

Thank you!

Hi Iriosmo,

(1) \(\displaystyle f(x)=2(x+2)^2-1\)

The standard (vertex) form of the equation of a parabola with vertex (h, k) and axis of symmetry x = h is \(\displaystyle f(x)=a(x-h)^2+k\)

This equation is already in standard form. So,

(a) Vertex (-2, -1)

(b) Since a > 0, k is the minimum value of the related function and the parabola opens upward. You might want to find the zeros and y-intercept to sketch a better graph.

(c) The axis of symmetry is x = -2

(d) Domain = {all real numbers}

Range = {y | y >= -1}

Your second function is in the general form \(\displaystyle f(x)=ax^2+bx+c\).

In this form, the y-intercept is c. The x-coordinate of the vertex is \(\displaystyle \frac{-b}{2a}\), and the axis of symmetry is \(\displaystyle x=\frac{-b}{2a}\).

To find the y-coordinate of the vertex, first find the x-coordinate as described above; then substitute that value into the original funcition to find f(x) or y.

(2) \(\displaystyle f(x)=2x^2+4x-3\)


I suggest you use a technique called 'completing the square' to put it into vertex form like the first one was in. Then, you can follow the steps above to answer the questions about this one. Let us know if you have trouble completing the square. You might take a look here if you don't remember: http://www.purplemath.com/modules/sqrvertx.htm

 

[tutor_joel]

to find the x intercepts, set the y value to zero and solve for x.

the domain is all values of x and the range is all values of y. In a parabola as these are, the domain is nearly always \(\displaystyle \pm\infty\), while range is the value of y from the vertex up or down to infinity, depending which way the parabola opens up. THis will be apparent once you sketch the graph.

 

[lriosmo]

Thank you all for your help! I can see I will be a frequent visitor here.