Combining evaluation, graphing and solving

We can use our understanding of functions and our knowledge of the shapes of curves to evaluate some of the key coordinates on graphs.

Intercepts

We usually want to label the intercepts of our graphs when we make a sketch. The y-intercept will have a zero as its first (x) coordinate: (0, ), and any x-intercept will have a zero as its second (y) coordinate: ( ,0). To decide on the missing coordinates we can think in the following way:

Intercepts
The coordinates of the y-intercept are   (0, the y-value corresponding to x=0)
The coordinates of any x-intercept(s) are   (any x-value making y=0, 0)
Example
Fill in the missing coordinates:
graph of y=x^3+1 with coordinates of x and y intercepts indicated as missing
To solve this problem we first fill in what we know without any calculations:
graph of y=x^3+1 with coordinates ( ,0) and (0, ) marked at intercepts
The y-intercept is
(0, (0)^3 + 1) = (0, 1)
and the x-intercept will be
(solution of x^3+1=0, 0)
The value x=-1 will make x^3+1=0, so the x-intercept is (-1,0):
graph of y=x^3+1 with the coordinates (-1,0) and (0,1) marked at intercepts

Intersections

When two graphs intersect, the point of intersection lies on both graphs, so the coordinates of the point of intersection will satisfy both of the equations which led to thses graphs. We need to solve the equations simultaneously, to find the coordinates of the intersection.

Examples
  1. Fill in the missing coordinates:
    graph of y=x^2-1 and y=x+1, with points of intersection marked
    We solve the equations simultaneously by setting them equal to one another:
    x^2-1=x+1, so x^2-x-2=0, or (x-2)(x+1)=0, and x=2 or x=-1
    Thus the x-coordinates of the two points of intersection are x=2 and x=-1. To get the corresponding y-coordinates we use either equation to find y(-1)=0 and y(2)=3. The graphs intersect at (-1,0) and (2,3):
    graph of y=x^2-1 and y=x+1, with coordinates (-1,0) and (2,3) of intersections marked
  2. Fill in the missing coordinates:
    graph of y=sin(x) with first maximum and x-intercept at the end of one period marked
    The sine function completes one full cycle between x=0 and x=2pi, so the right-hand point has coordinates (2pi,0). The graph reaches its highest value one quarter of the way through the cycle, at x=pi/2. Since sin(pi/2) = 1, the coordinates of the left-hand point are (pi/2,1):
    graph of y=sin(x) with coordinates (2pi,0) and (pi/2,1) marked
    (See also section 14, especially topic 1, common graphs to memorize and topic 2, key function values, for more information on the sine function.)
Practice

Be sure you've gone through each step in the examples above before doing these. Once you've worked them until you're sure that you understand them, go on to the next section. There are more problems of this type in the section test at the end of the section.

Note that you can get new practice problems by clicking the "Refresh" button at the bottom of the practice set.

  1. The graph of the line y = -6 x + 5 is shown below. Find the coordinates of the points P and Q:
    graph of line with negative slope, P at the intersection of the line with the y-axis, Q at that with the x-axis
    P = (, )
    Q = (, )
    (Enter answers, then click: . Answer message: )
  2. The graph of the parabola
    y   =   c x ^2   +   d x   -   e

    is shown below. Find the coordinates of the points P, Q and R.
    graph of parabola with y intercept Q and P and R at the intersections of the parabola with the x-axis
    P = (, )
    Q = (, )
    R = (, )
    (Enter answers, then click: . Answer message: )
  3. The graphs of the parabola
    y   =   - x ^2   +   f x   +   g

    and line
    y   =   h x   +   k

    are shown in the graph below. Find the indicated points P and Q.
    graph of parabola and line, with P and Q being the intersections of the two
    P = (, )
    Q = (, )
    (Enter answers, then click: . Answer message: )
For more practice, click: .
next page
section test

precal: 11.1 - evaluation,graphs & solving
page created: Sat Apr 19 15:34:30 2014
Comments to math-itc(at)umich(dot)edu
©2003-2007 Gavin LaRose, Pat Shure / University of Michigan Math Dept. / Regents of the University of Michigan