Page 123 - Algebra

Step 3: Find the x-intercept (point where y is 0) by replacing f(x) with 0. ax2 +bx+c=0
Step 4: Solve the quadratic equation for x, you get two values x1 and x2. The x-intercept are (x1, 0) and (x2, 0)
Step 5: Similarly, find the y-intercept (the point where x is 0) by replacing x with 0. The y-intercept is (0, y1)
Step 6: Plot the intercept on the graphs.
Worked Example
Plot the graph of the following equation
f(x) = x2 – 8x + 12
Solution:
Here a > 0, so it’s an upward parabola. a = 1, b = – 8, c = 12
Vertex ((–
4))
f(
The vertex is (4, – 4) Put f(x) = 0
y = 12
= (–
−8 2
, f(
𝑏𝑏
, f(– ))
2𝑎 2𝑎
= (4, f(4))
4)=42 –8(4)+12=16–32+12=–4
x2 – 8x + 12 = 0
x2 – 6x – 2x + 12 = 0
x(x – 6) – 2(x – 6) = 0
(x – 6) (x –2) = 0
x = 6 or 2
x-intercept are (2, 0) and (6, 0)
Put x = 0,
y = x2 – 8x + 12
y –intercept is (0, 12)
Now, plot these points on the graph.
y
x A(1, x
’
4)
A(4,
-4)
y ’
Page 122 of 177
Algebra I & II