a Let's complete the square. To do that, we will consider the variable terms on the function's right-hand side.
f(x)= x^2+6x+15
To visualize how we complete the square, we draw a generic rectangle where the upper left corner has an area of x^2 and the adjacent rectangles each have an area that is half of 6x.
Since the upper left corner has an area of x^2, it must be a square with a side length of x. This allows us to factor the adjacent rectangle's areas to 3* x. With this information, we can also determine the area of the lower right rectangle which completes the square.
As we can see, we need to add 3^2 to complete the square. To keep the equation true, this term must be added to both sides.
f(x)+ 3^2= x^2+2(3x)+ 3^2+15
Let's isolate f(x) and simplify.
When a quadratic function is written in graphing form, we can identify its vertex.
Graphing Form:& f(x)=a(x- h)^2+ k
Vertex:& ( h, k)
By rewriting our function so that it matches the graphing form exactly, we can find its vertex.
Function:& f(x)=(x-( -3))^2+ 6
Vertex:& ( -3, 6)
The function has its vertex in (-3,6) which means the axis of symmetry goes through x=-3.
b Like in Part A, we will complete the square by adding half the coefficient to x to both sides of the equation.
Examining the function, we see that it matches the graphing form of a parabola exactly. Let's identify the vertex.
Function:& y=(x- 2)^2+ 5
Vertex:& ( 2, 5)
The function has a vertex in (2,5) which means the axis of symmetry goes through x=2.
c In this case, the right-hand side is missing a constant. This means we can solve for the x-intercepts by factoring out x and using the Zero Product Property.
The x-intercepts are at (0,0) and (8,0). 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 (4,- 16). When we know the vertex, we can write the function in graphing form where ( h, k) shows the parabola's vertex.
Graphing Form:& f(x)=a(x- h)^2+ k
Function:& f(x)=1(x- 4)^2+( -16)
This simplifies to f(x)=(x-4)^2-16.
d In order to pick a method, we should first consider if the function actually has x-intercept(s). Examining the equation, we can conclude a couple of things that will help us decide this.
Its constant is -2 which means it intercepts the y-axis at (0,-2).
It has a positive x^2-term which means the parabola opens upward.
These two facts must mean that it does have two x-intercepts as the curve must dip below the x-axis in order to intercept the y-axis at (0,-2). Therefore, let's solve for the x-intercepts using the Quadratic Formula.
The vertex is at (-3.5,- 14.25). 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-( - 3.5))^2+( -14.25)
This simplifies to y=(x+3.5)^2-14.25.
