Example Solution: y=x^2-10x+27 and y=- x^2+10x-15
Explanation: See solution.
Practice makes perfect
We want to find two parabolas that go through the points (3,6) and (7,6). Let's graph these points.
Notice that the line x=5 is in the middle of these two points, so it can serve as the axis of symmetry for our two parabolas. Recall the formula for the axis of symmetry.
x=- b/2a ⇒ 5=- b/2a
Finding the First Quadratic
To find our first equation we can choose values for a and b that satisfy the equation for the axis of symmetry. Let a=1 and solve for b.
We have found that a=1 and b=- 10 allows us to have an axis of symmetry at x=5.
y=ax^2+bx+c ⇒ y=x^2-10x+c
We will find a c-value that makes the parabola pass through the points (3,6) and (7,6). Since parabolas are symmetric, if the parabola passes through one of the points it will also pass through the other. Substitute (3,6) for (x,y) and solve for c.
Let's substitute this into our equation.
y=x^2-10x+27
This is our first equation.
Finding the Second Quadratic
Now we will make another equation that passes through the points (3,6) and (7,6). This parabola will have the same axis of symmetry.
5=- b/2a
The a and b values we use can be the negative of what we used for our first quadratic, so we will use a=- 1 and b=10.
y=ax^2+bx+c ⇒ y=- x^2+10x+c
Now we will find a c-value that allows the parabola to go through the point (3,6). Substitute (3,6) for (x,y).
