a All parabolas are symmetric about their vertex. What this means is if two points have the same y-coordinate, such as the x-intercepts, they are equidistant from the parabola's line of symmetry. Notice that the function's right-hand side is written in factored form. Therefore, we can find its x-intercept(s) by using the Zero product Property.
The x-intercepts are at (3,0) and (11,0). Since the x-coordinates are equidistant from the line of symmetry, which runs through the vertex, we can find the vertex x-coordinate by averaging the x-intercepts.
From the diagram we see that the vertex has a y-coordinate of -16. However, if it wasn't easy to see it from the diagram, we could always calculate it by substituting the vertex x-coordinate into the function.
The vertex is at (7,- 16). When we know the vertex, we can write the function in graphing form where (h,k) shows the parabolas vertex. Notice that the parentheses isn't preceded by a coefficient which means a=1.
Graphing Form:& y=a(x-h)^2+k
Function:& y=1(x-7)^2+(-16)
We can simplify this to y=(x-7)^2-16.
b Like in Part A, we can find the x-intercepts by setting y equal to 0 and solving for x with the Zero Product Property.
Again, to find the vertex x-coordinate, we will average the x-intercepts.
From the diagram we see that the vertex has a y-coordinate of -16. However, if it wasn't easy to see it from the diagram, we could always calculate it by substituting the vertex x-coordinate into the function.
The vertex is at (2,- 16). When we know the vertex, we can write the function in graphing form.
Graphing Form:& y=a(x-h)^2+k
Function:& y=1(x-2)^2+(-16)
We can simplify this to y=(x-2)^2-16.
c This function is not written in factored form. In the way it is written, we are best off solving for the x-intercepts using the Quadratic Formula.
The vertex is at (7,- 9). Like in Parts A and B, we can use this to find the graphing form of the equation.
Graphing Form:& y=a(x-h)^2+k
Function:& y=1(x-7)^2+(-9)
We can simplify this to y=(x-7)^2-9.
d By setting y equal to 0, we can solve the for x-intercepts by performing inverse operations until x is isolated.
The x-intercepts are at x=1 and x=3. By averaging these values, we can find the vertex x-coordinate.
From the diagram we can see that the vertex has a y-coordinate of -1. Notice that the function is already written in graphing form which confirms this.
Graphing Form:& y=(x-2)^2-1
Vertex:& (2,-1)
