Exercise 13 Page 528

As we can see, the graph crosses the x-axis at the points (- 6, 0) and (- 4,0). Thus, the x-intercepts are - 6 and - 4. To write the equation of the parabola with x-intercepts - 6 and - 4, we will use the factored form of a quadratic function. y=a(x+b)(x+c) ⇕ y=a(x-(- b))(x-( - c)) In this form, - b and - c are the intercepts. With this information we can partially write our equation. y=a(x-(- 6))(x- ( - 4)) ⇕ y=a(x+6)(x+4) We can see in the graph that the parabola opens upwards. Thus, a will be a positive number. Since a does not have any effect on the x-intercepts, we can choose any positive value. For simplicity we will let a=1. y=1(x+6)(x+4) ⇕ y=(x+6)(x+4) We found a possible equation for the given parabola. Please note that this is just one example of a quadratic function that has the given x-intercepts.

As we can see, the lowest point of the graph lies on the x-axis. Thus, there is only one x-intercept, which is also a vertex. Since our function has the x-intercept at (- 1,0), we know that the vertex of the parabola is (- 1,0). This time, to write the equation we will use the vertex form of a quadratic function. y= a(x- h)^2+k The vertex of this parabola has coordinates ( - 1, ). This means that we have h= - 1 and k= . We can use these values to partially write our equation. y= a(x-( -1))^2+ ⇕ y= a(x+1)^2 We can see in the graph that the parabola opens upwards. Thus, a will be a positive number. Since a does not have any effect on the x-intercepts, we can choose any positive value. For simplicity we will let a= 1. y= 1(x+1)^2 ⇕ y=(x+1)^2 We found a possible equation for the given parabola. Please note that this is just one example of a quadratic function that has the given x-intercept.