a Since the function is given in standard form, we can easily find the y-intercept — it is the value of the constant term in the equation.
y = x^2 + 6x + 15
Next, to write the function in the graphing form 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 area to 3* x. With this information we can also determine the area of the lower right rectangle that 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 at (-3,6).
b Like in Part A, the function is given in the standard form so the y-intercept is the value of the constant term in the equation.
f(x)=x^2-3x+( -9)
Next, to write the function in the graphing form let's complete the square. To do that we will consider the variable terms on the function's right-hand side.
f(x)= x^2-3x-9
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 area to -1.5* x. With this information we can also determine the area of the lower right rectangle that completes the square.
As we can see, we need to add (-1.5)^2 to complete the square. To keep the equation true, this term must be added to both sides.
f(x)+ (-1.5)^2= x^2+2(-1.5x)+ (-1.5)^2-9
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-1.5)^2+( -11.25)
Vertex:& (1.5, -11.25)
The function has its vertex at (1.5,-11.25).
