Q:  Consider the equation:

y= x²-4x+3

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= 0²-4(0)+3

y = 3

The y-interepts occur when y = 3 and is the point (0, 3)