Q: Consider the equation:
1. Find the vertex.
2. It opens _____.
3. Find the axis of symmetry.
4. Find the minimum.
5. Find the discriminant.
6. It has ___ real solutions.
7. Find the x-intercept(s), if none, say so.
8. Find the y-intercept.
A: We start by recognizing: y= x²-4x+3 means a = 1, b = -4, c = 3:
(1) The x-coordinate of the vertex can be found by computing x = -b/(2a):
x = -(-4)/(2*1) = 4/2 = 2
To find the y-coordinate of the vertex, plug x=2 (found above) into the original equation:
y = 2² – 4(2) + 3 = -1
The vertex is the point (2, -1)
(2) The parabola opens up. We know this since “a” is positive. (a = 1, remember)?
(3) The axis of symmetry is the line you could draw right down the middle of it. This is a vertical line and is located at x = -b/(2a). Remember we just calculated this? x=2 is the axis of symmetry.
(4) Find the minimum. The minimum occurs at the vertex. It occurs when x = 2 and is -1.
(5) Find the discriminant…. The discriminant = b² -4ac (this is the stuff that is inside the square root in the quadratic formula).
So, b² – 4ac = (-4)² – 4(1)(3) = 4
(6) It has two real solution.
How did I know this? The discriminant told me. If the discriminant is positive, there are 2 real solutions. If the discriminant equals 0, there is 1 real solution. If the discriminant is negative, there are no real solutions.
(7) Find the x-intercepts…. This is where the function cross the x-axis and it occurs when y=0 (x-intercepts are considered to be “the solutions” from part 6… therefore, we know there are two of them). So, plug in 0 for y to get:
0 = x² – 4x + 3
Since it is a quadratic, we can use the quadratic formula or factoring to solve. I will factor:
0 = (x -1)(x – 3)
Therefore, x = 1 and x = 3 are the x-intercepts. The x-intercepts occur at the points: (1, 0) and (3, 0).
(8) Similar to part (7), the y-intercept occurs when x=0. Plug in o for x to get:
y = 3
The y-interepts occur when y = 3 and is the point (0, 3)