Answer is ( -1, 6) and (2, 3) the points where the parabola and the linear line intersect
First solve -x + 5
y = -x + 5
Sub 0 for y and solve for x
0 = -x + 5
add x to each side to solve
0 + x = -x + x + 5
Simplify
x = 5
y = -x + 5
Sub 0 for x and solve for y
y = 0 + 5
y = 5
So we have (0, 5) and ( 5, 0) to draw your straight line
Now your parabola
X^2 - 2x + 3
Find the x of your vertex
- b/2a
- -2/ 2*1
2/2 = 1
X = 1
Sub x value of 1 into equation to find y
1^2 -(2*1 ) + 3
1 -2 +3
y = 2
So your vertex of the parabola is ( 1, 2)
Now we can find another point on the parabola to draw it
Since x vertex is 1, let’s try -1 to sub into the equation X^2 - 2x + 3
-1^2 (-2*-1) + 3
1 +2 +3
y = 6
So we can plot point ( -1, 6) on the graph
We know the axis of symmetry is x=1 so to find the other point of symmetry we see -1 is 2 places to the left to ( -1, 6) so we go 2 places to the right of x=1 + 2 = 3
Our symmetry point is (3, 6)
Draw your parabola and see where the parabola and the linear line intersect
Graph is attached