A step by step procedure on how to graph a quadratic equation with only 1 solution (discriminant is 0) and no real solutions (complex and discriminant is negative) ? Please provide an example equation and solve through it...

2 Answers

Answer:

Here you go:

Explanation:

(All work is done with quadratic converted to vertex form. Graphing tends to be easier when converted to vertex form)

There are two ways to solve this problem, and they both work no matter what your discriminant is.

Let's try the equation #f(x) = (x- 6)^2 + 5#, whose graph looks like this:
graph{(x-6)^2 +5 [-1, 10, -1, 10]} *

First Method: Counting Points
Caution: This method is not generally encouraged by math teachers but is a general rule for quadratics.

1.) Because our answer is in vertex form, we already have one point: the vertex. Go ahead and graph that first. In the case of this function, the vertex is at the point #(6, 5)# .

2.) Now that you have the vertex, start from that point and go up one point and over one point (in both directions) and plot those points.

3.) When you finish with that, move over one point and up three from the previous two you graphed.

5.) Do this as many times as needed, but every time you go to plot your y-value, make sure that you go up to more than you did the previous time (so if you went up one, go up three. If you went up three, go up five. If you went up five, go up seven).

6.) This method is supported by the graph of the function #f(x) = (x-6) ^ 2 +5# :

X | Y
4 | 9
5 | 6
6 | 5
7 | 6
8 | 9

See how the y-values are only one higher than they are at the vertex when the function is evaluated at #f(5)# and #f(7)#? And do you see how the y values at #f(4)# and #f(8)# are 3 higher than they were previously? This method works for any quadratic as long as it doesn't have a vertical or horizontal stretch. If it is stretched, you have to adjust the rules accordingly.

Second Method: Calculating Y-Values
This method is supported by math teachers

1.) Again, start by graphing the vertex, which is at point #(6,5)#.

2.) Start plugging in values for #x# and listing them out in a table of coordinates (See above for coordinates of function #f(x) = (x-6)^2 +5#)

3.) Graph the y-value you find for whatever value you plugged in for #x#.

4.) Example:

#f(4) = 9#
#f(5) = 6#
#f(6) = 5#
#f(7) = 6#
#f(8) = 9#

Jun 29, 2017

Answer:

Hope this is what you were looking for! Didn't average any roots but it seems to answer your question about not having any!

Explanation:

Let's say we want to graph the function #f(x) = x^2 - 12x + 41#, whose graph looks like this:
graph{x^2 -12x +41 [-5.15, 18.78, -0.74, 11.22]}

But how do we get here from the standard form of the equation? First, let's start by finding the vertex.

Now the first step in finding the vertex is finding the axis of symmetry (AOS), which is the verticle line that completely bisects the function. The line's equation follows the pattern #x=h#, and because the vertex is the middlemost point of the parabola, the axis of symmetry will be the #x#-coordinate of our vertex.

To find the AOS, we simply have to use the equation #h=-b/(2a)#, where h is the location of the AOS and the #x#-coordinate of the vertex. (#h# is generally used because when writing the coordinates of a vertex, we usually use the form #(h,k)#, as seen when writing a quadratic in standard form, which follows the pattern #y=(x-h)^2 + k#)

So, to find the AOS of this graph we simply plug-in #1# and #12# for #a# and #-b# (respectively), and we find that #(-(-12))/(2(1)) = 6#, meaning that our AOS is #x=6# and the #x#-coordinate of our vertex is #(6,k)#.

Well, that's great and all, but now how do we find #k#? Simple! We just evaluate the equation at #f(h)#! This works because h is the #x#-coordinate of our vertex. When dealing with functions, we know that all we are doing is plugging in #x#-values to find out what our matching #y#-values are. So, if we have an #x#-value of 6, all we have to do is plug in this value for x, and we will find our matching #y#-value! So let's go ahead and do that:

#f(6) = (6)^2 -12(6) + 41 = 5#

Awesome! Now we know that our #k#-value is #5#, which gives us our vertex at #(6,5)#. Since we are graphing in standard form, go ahead and plot this point on your graph.

Now to continue. So we have our vertex, but how do we figure out the rest of our points? Simple! We plug-in values for #x# to find our values for #y#, just as we did when we found our vertex.

Let's go ahead and do that:

#f(3) = (3)^2 - 12(3) + 41 = 14#
#f(4) = (4)^2 -12(4) + 41 = 9#
#f(5) = (5)^2 -12(5) + 41 = 6#
#f(7) = (7)^2 -12(7) + 41 = 6#
#f(8) = (8)^2 -12(8) + 41 = 9#
#f(9) = (9)^2 -12(9) + 41 = 14#

Vuala! Now we can graph these points, and thus, our function has been graphed as well! The graph of this function above will prove my work and I hope this helps!