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Find the vertex and the axis of symmetry of the parabola.
Graph:
Solutions: No solutions.
To solve the system of equations by graphing, we will draw the graph of the two quadratic functions on the same coordinate grid. Let's start with 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 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.
We use the x-coordinate of the vertex to find its y-coordinate by substituting it into the given equation.
x= - 1/2
(a/b)^m=a^m/b^m
a+(- b)=a-b
a/b=a * 2/b * 2
a = 4* a/4
Add and subtract fractions
The y-coordinate of the vertex is 114. Therefore, the vertex is at the point (- 12, 114). With this, we also know that the axis of symmetry of the parabola is the line x=- 12. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
| 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.
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.
a= - 1, b= - 3
a(- b)=- a * b
- a/- b= a/b
Put minus sign in front of fraction
We use the x-coordinate of the vertex to find its y-coordinate by substituting it into the given equation.
x= - 3/2
(a/b)^m=a^m/b^m
- a(- b)=a* b
a*b/c= a* b/c
a/b=a * 2/b * 2
a = 4* a/4
Add and subtract fractions
The y-coordinate of the vertex is - 114. Therefore, the vertex is at the point (- 32,- 114). With this, we also know that the axis of symmetry of the parabola is the line x=- 32. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
| 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.
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.