This expression is called the Discriminant, also shown as Δ.
It is equal to b² - 4ac. This is a very important part of the quadratic formula as it determines whether x will have two values, one repeated value or no real values. Here are a few examples.
a) x² - 2x - 1. a is equal to 1 since 1x² = x². b = -2, c = -1
The discriminant will be (-2)² - 4×1×-1 = 4 + 4 = 8.
Since Δ > 0, there are two x values. Graphed, the parabola sinks below the x axis.
b) x². a = 1, b = 0 (0x = 0), c = 0
The discriminant will be 0² - 4×1×0 = 0 - 0 = 0.
Since Δ = 0, there is only one x value. Graphed, the parabola touches the x axis at only one point.
c) x² + 1. a = 1, c = 1.
The discriminant will be 0² - 4×1×1 = 0 - 4 = -4
Since Δ < 0, there are no real x values. Graphed, the parabola floats above the x axis.
Hope this helps!
1.
f(x) = x
g(x) = x + 3
For each x value you use in f(x) and g(x), the corresponding y value in g(x) is always 3 more than in f(x). If each y-coordinate is 3 more, that means the graph is shifted 3 units up. The graph of f(x) is shifted 3 units up to create the graph of g(x).
2.
The first term is -0.5.
Then each term goes up by 0.25.
1st term: -0.5
2nd term: -0.5 + 0.25 = -0.5 + 0.25(1)
3rd term: -0.5 + 0.25 + 0.25 = -0.5 + 0.25(2)
4th term: -0.5 + 0.25 + 0.25 + 0.25 = -0.5 + 0.25(3)
nth term: -0.5 + 0.25(n - 1)
-0.5 + 0.25(n - 1) = -0.5 + 0.25n - 0.25 = -0.75 + 0.25n = 0.25n - 0.75
Answer is C.
This would be a great question to just plug into a calculator. I'm sure you will find the answer quite easily if you use a calculator.
Evaluate the function
g(x) = 2x2 + 3x – 5 for the input values -2, 0, and 3.
This is tedious math work but necessary to sharpen your skills.
Let x = -2
g(-2) = 2(-2)^2 + 3(-2) – 5
g(-2) = 2(4) - 6 - 5
g(-2) = 8 - 11
g(-2) = -3
Now let x = 0 and repeat the process.
g(0) = 2(0)^2 + 3(0) - 5
g(0) = 0 + 0 - 5
g(0) = -5
Lastly, let x = 3.
g(3) = 2(3)^2 + 3(3) - 5
g(3) = 2(9) + 9 - 5
g(3) = 18 + 9 - 5
g(3) = 27 - 5
g(3) = 22
Did you follow through each step?
