Finally we can express our function in graphing form, f(x)=a(x-h)^2+k, where a, h, and k are either positive or negative constants.
f(x)=(x+2)^2+1
⇕
f(x)=1(x-(-2))^2+1
Now, we can state the vertex of the parabola. It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the graphing form to our equation.
General Formula:y=& a(x- h )^2 +k
Equation:y=& 1(x-( -2))^2+1
We can see that a= 1, h= - 2, and k=1.
The vertex of a quadratic function written in graphing form is the point ( h,k). For this exercise, we have h= - 2 and k=1. Therefore, the vertex of the given equation is ( - 2,1).
b We want to rewrite the given function in the graphing form by completing the square.
f(x)=x^2-7xIn a quadratic expression, b is the linear coefficient. For the function above, we have that b=- 7. Let's now calculate ( b2 )^2.
Finally we can express our function in graphing form, f(x)=a(x-h)^2+k, where a, h, and k are either positive or negative constants.
f(x)=(x-3.5)^2-12.25
⇕
f(x)=1(x-3.5)^2+(- 12.25)
Now we can state the vertex of the parabola. It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula of the graphing form to our equation.
General Formula:y=& a(x- h)^2 + k
Equation:y=& 1(x- 3.5)^2+(- 12.25)
We can see that a= 1, h= 3.5, and k=- 12.25.
The vertex of a quadratic function written in graphing form is the point ( h,k). For this exercise, we have h= 3.5 and k=- 12.25. Therefore, the vertex of the given equation is ( 3.5,- 12.25).
c Before we determine the maximum or minimum recall that, if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards.
In both given functions we have a= 1, which is greater than 0. Thus, the parabolas open upwards and in both of them the vertex represents a minimum value.
