To solve the by graphing, we will draw the graph of the two on the same coordinate grid. Let's start with the first .
Graphing the First Parabola
To graph the parabola, we first need to identify a, b, and c.
3=y-x^2-x ⇔ y= 1x^2+ 1x+ 3
For this equation we have that a= 1, b= 1, and c= 3. Now, we can find the 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.
We use the x-coordinate of the vertex to find its y-coordinate by substituting it into the given equation.
y=x^2+x+3
y=( - 1/2)^2+( - 1/2)+3
y=1/4+(- 1/2)+3
y=1/4-1/2+3
y=1/4-2/4+3
y=1/4-2/4+12/4
y=11/4
| x
|
x^2+x+3
|
y=x^2+x+3
|
| 1
|
1^2+ 1+3
|
5
|
| - 2
|
( - 2)^2+( - 2)+3
|
5
|
Both ( 1, 5) and ( - 2, 5) are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Graphing the Second Parabola
Now, we need to repeat the procedure to draw the second parabola. Let's start by identifying a, b, and c.
y=- x^2-3x-5 ⇕ y= - 1x^2+( - 3)x+( - 5)
For this equation we have that a= - 1, b= - 3, and c= - 5. Now, we can find the vertex using its formula. Let's find the x-coordinate of the vertex.
- b/2a
- - 3/2( - 1)
- - 3/- 2
- 3/2
-3/2
y=- x^2-3x-5
y=- ( - 3/2)^2-3( - 3/2)-5
y=- 9/4-3(- 3/2)-5
y=- 9/4+3(3/2)-5
y=- 9/4+9/2-5
y=- 9/4+18/4-5
y=- 9/4+18/4-20/4
y=- 11/4
| x
|
- x^2-3x-5
|
y=- x^2-3x-5
|
| 0
|
- 0^2-3( 0)-5
|
- 5
|
| - 3
|
- ( - 3)^2-3( - 2)-5
|
- 5
|
Both ( 0, - 5) and ( - 3, - 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 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.
