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To find the number of points of intersections, draw the graphs of both functions on the same coordinate grid.
Two points of intersections
We want to determine the number of points of intersections of the given system of equations. Note that the first equation of the system is quadratic equation and the second one is linear equation. y=- (x-1)^2+3 & (I) 2x+3y-6=0 & (II) In other words, we want to find the number of points in which the graphs of the given equations — a parabola and a line — intersect. When it comes to a line and a parabola, we have three possibilities.
To graph the parabola, let's recall the graphing form of a quadratic function. y=a(x- h)^2+k In this form the vertex of the parabola is the point ( h,k), and the axis of symmetry is the vertical line x= h. Now consider the given function. y=- (x-1)^2+3 ⇕ y=- 1(x- 1)^2 + 3 We can see that h= 1 and that k=3. Therefore, the vertex is ( 1,3), and the axis of symmetry is x= 1. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.
x | - (x-1)^2+3 | y=- (x-1)^2+3 |
---|---|---|
- 1 | - ( - 1-1)^2+3 | - 1 |
0 | - ( 0-1)^2+3 | 2 |
2 | - ( 2-1)^2+3 | 2 |
3 | - ( 3-1)^2+3 | - 1 |
Let's now draw the axis of symmetry x=1 and the parabola that connects the obtained points and the vertex.
LHS-2x=RHS-2x
LHS+6=RHS+6
.LHS /3.=.RHS /3.
Put minus sign in front of fraction
Finally, we can determine the number of points of intersections.
As we can see, the graphs of the given functions have two points of intersection.