Question 134725This question is from textbook College Algebra
: I am completely lost on how to do these. Any help would be great.
It says:
Sketch the graph of the following. Label vertex, xand yintercepts.
f(x)=x^2x12
Vertex=( , )
xintercepts=( , );( , )
yintercept=( , )
This question is from textbook College Algebra Found 3 solutions by cbcrawford978, ankor@dixienet.com, solver91311: Answer by cbcrawford978(7) (Show Source): Answer by ankor@dixienet.com(22719) (Show Source):
You can
put this solution on YOUR website! I am completely lost on how to do these. Any help would be great.
It says:
Sketch the graph of the following. Label vertex, xand yintercepts.
f(x) = x^2 – x – 12
Here’s how you can sketch a graph
f(x) = y so we have:
y = x^2 – x – 12
:
We will find the values of y by substituting 5 to +5 in the equation
A table of these value will be:
x  y
——
5 +18
4 +8
3 0; when y = 0, this is one of the x intercepts
2 6
1 10
0 12; notice when x = 0, the value of y is 12, this is the y intercept
+1 12; it has reached a minimum and started back up again
+2 10
+3 6
+4  0; this is the 2nd x intercept
+5 +8
Plot these values, it should look like this:
:
Vertex=( , )
To find the vertex, first find the axis of symmetry. This can be found using the formula x = b/(2a)
In this equation a=1; b=1
x =
x = or .5
Find the vertex by substituting 1/2 for x in the original equation:
y = .5^2 – .5 – 12
y= .25 – .5 – 12
y = 12.25
The x/y values for the vertex is: .5, 12.25, this is the minimum as you can see by the graph
:
xintercepts=( , );( ,
x intercepts are the points which the graph crosses the x axis (y=0)
you can find them by solving the equation
x^2 – x – 12 = 0
Factors to
(x4)(x+3) = 0
x = +4
x = 3; the x intercepts
:
yintercept=( , )
As we said the y intercept occurs when x = 0, you see when substitute 0 for x in the equation; y = 12
:
Did this help you? Do you have any questions?
Answer by solver91311(24713) (Show Source):
You can
put this solution on YOUR website! Your function is in the form where , , and
The vertex is located at the point (,). So, calculate to get the xcoordinate, then evaluate to get the ycoordinate. Plot this point on your graph.
The xintercepts are at (p,0) and (q,0) where and are the solutions to f(x)=0. So solve to determine p and q, then plot these two points.
The yintercept is at (0,f(0)). Evaluate f(0) by substituting 0 for x in . Plot this point.
Multiply the xcoordinate of the vertex by 2. Symmetry of the curve about the vertical line means that there will be another point on the curve at (,), i.e. 2 times the xcoordinate of the vertex and the ycoordinate of the yintercept. Plot this point.
Draw a smooth curve through the plotted points. Done.
It should look something like this:

