They all have the same graph. By rewriting f(x) and g(x) in standard form, we obtain that they both have the same equation as h(x).
Practice makes perfect
We need to compare the graphs of the functions below.
f(x) &= (x+3)(x+1)
g(x) &= (x+2)^2 - 1
h(x) &= x^2 + 4x + 3
Let's draw the graph of each quadratic function separately.
Graphing f(x)
Let's recall the intercept form of a quadratic function.
y=a(x-p)(x-q)
Here, p and q are the x-intercepts of the parabola. Let's now consider the first function.
f(x)=(x+3)(x+1)
⇕
f(x)= 1(x-( - 3))(x-( - 1))
We see that the x-intercepts are -3 and -1. Since a = 1 is greater than 0, the parabola opens upward. Now let's find its vertex, knowing that its x-coordinate is halfway through p and q.
x=p+q/2
y=f(p+q/2)
Vertex
x = -3+(-1)/2 ⇕ x= -2
y =f( -2) ⇕ y=( - 2+3)( - 2+1) ⇕ y= -1
( -2, -1)
We will draw the parabola that connects the x-intercepts with the vertex.
Graphing g(x)
In contrast to the first function, the second one is given in vertex form.
y=a(x-h)^2+k ⟶ Vertex: (h,k)
Let's identify the values of a, h, and k in the given function.
g(x)=(x+2)^2-1
⇕
g(x)= 1(x-( - 2))^2 + ( - 1)
We see that a= 1, h= -2, and k= -1, which means the vertex is ( -2, -1). Let's find the x-intercepts. To do so, we will solve the equation g(x)=0.
We will find one of the x-intercepts by using the positive sign, and the other by using the negative sign.
c|c
x_1=- 2+1 & x_2=- 2-1
⇕ & ⇕
x_1=- 1 & x_2=- 3
Let's now draw the parabola that connects the x-intercepts with the vertex.
Graphing h(x)
Finally, the function h(x) = x^2 + 4x + 3 is given in standard form y=ax^2+bx+c. From this we can get the vertex of the parabola.
Function
Vertex
h(x)= ax^2+ bx+ c
x=-b/2a
y=g(-b/2a)
h(x) = 1x^2 + 4x + 3
x=-4/2( 1) ⇕ x=-2
y=h(-2) ⇕ y= -1
Therefore, the vertex is (-2, -1). Next, by using the Quadratic Formula we will find the x-intercepts.
The x-intercepts for this equation are x= -4±22. Let's separate them by using the positive and negative signs.
x=-4± 2/2
x=-4 + 2/2
x=-4 - 2/2
x=-2/2
x=-6/2
x=-1
x=-3
The parabola intercepts the x-axis at -3 and -1.
Comparing the Graphs
After plotting the three functions separately, we see that they are all the same parabola. To see why it happens, we can rewrite the functions f(x) and g(x) in standard form.
