5. Draw the graph of y = f(x) = x2 + 2x - 8.


x2 + 2x - 8 > 0


Before we draw the graph it would help to answer questions a), b), and c).


a) What is the y-intercept?


The y-intercept is the point on the graph whose x-coordinate is 0. So let x = 0 in the equation, and we get


y = (0)2 + 2(0) - 8


= 0 + 0 - 8


= -8.


This illustrates that the y-intercept is always the constant term in a polynomial.


b) What are the roots?


The roots are another name for the x-intercepts. They are the points on the graph whose y-coordinates are 0. So let y = 0 in the equation and we get


0 = x2 + 2x - 8


This gives us a quadratic equation in x. We are fortunate enough to already have a 0 on one side so we are ready to factor.


0 = (x - 2)(x + 4)


Set the factors = 0.


x - 2 = 0 or x + 4 = 0


Solve for x


x = 2 or x = -4



c) What are the coordinates of the vertex?


If we use the formula for the x-coordinate of the vertex,



we get





x = -1


Now that we have the x coordinate of the vertex, we run this number through the function to find the y coordinate


y = f(-1) = (-1)2 + 2(-1) - 8


= 1 - 2 - 8


= -9


So the vertex is at (-1, -9)


It is a good idea to get all of this information before plotting points to draw the graph. When we plot points, since we now know that the x coordinate of the vertex is -1, we will want -1 to be in the middle of the x's that we plot. We will also want the roots, x = -4 and 2 to be in the interval containing the x's that we plot, and at least one point on the other sides of the roots from the vertex.



When we plot these points, we get



d) Write the equation in vertex form.


The equation is given to be


y = x2 + 2x - 8


Half of the linear coefficient is 1, and the square of 1 is 1, so we add and subtract 1


y = x2 + 2x + 1 - 1 - 8


This simplifies to


y = (x + 1)2 - 9


and we see the coordinates of the vertex in the equation.


e) Solve for x



In the process of drawing the graph we found that the roots were at -4 and 2. When we divide the real number line into the intervals between the roots, and check out an x in each interval.



Since equality is permitted, the endpoints of the intervals, the roots, being the x's which will give answers which are equal to 0 when substituted into the formula, will be solutions. So we draw square brackets around the endpoints.


The set of solutions is the set of x's which satisfy


In interval notation, this comes out to be


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