Notice that the first function is given in the vertex form. Therefore, we can identify the values of a, h and k. Then we will be able to find the vertex of the first parabola.
y=(x+2)^2-3
⇕
y= 1(x-( - 2))^2+( - 3)
For this equation we have that a= 1, h= - 2, and k= - 3. Therefore, we know that the vertex of the given function is ( - 2, - 3). With this, we also know that the axis of symmetry of the parabola is the line x=- 2. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
x
(x+2)^2-3
y=(x+2)^2-3
0
( 0+2)^2-3
1
- 4
( - 4+2)^2-3
1
Both ( 0, 1) and ( - 4, 1) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Graphing the Second Parabola
Let's now draw the second parabola. First we need to identify a, b, and c.
y=x^2+4x+5 ⇔ y= 1x^2+ 4x+ 5
For this equation we have that a= 1, b= 4, and c= 5. Now, we can find the vertex using its formula. To do this, we will need to think of y as a function of x, y=f(x).
Vertex of a Parabola: ( - b/2 a,f(- b/2 a) )
Let's find the x-coordinate of the vertex.
The y-coordinate of the vertex is 1. Therefore, the vertex is at the point (- 2,1). With this, we also know that the axis of symmetry of the parabola is the line x=- 2. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
x
x^2+4x+5
y=x^2+4x+5
0
0^2+4( 0)+5
5
- 4
( - 4)^2+4( - 4)+5
5
Both ( 0, 5) and ( - 4, 5) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Finding the Solutions
Finally, let's try to identify the coordinates of the points of intersection of the two parabolas.
As we can notice on the graph, the parabolas do not intersect. This means that the given system of equations does not have any solutions.
