Is the discriminant positive, negative, or 0 for each graph below? Explain your choices in one or two sentences for each graph.

Step-by-step explanation:
Ok so as you may the quadratic formula is: [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
And the discriminant is: [tex]b^2-4ac[/tex]
The reason this is so important, is because we can just calculate this and find how many solutions and what type of solutions a quadratic has.
Let's look at the three cases:
Positive:
If the discriminant is positive, there are two real solutions, since there will be a negative and positive square root solution, so there will be two distinct solutions
Negative:
If the discriminant is negative, there are two imaginary solutions, and what this really means is if you graph the equation, the quadratic never crosses the x-axis
Zero:
There is one real solution, since sqrt(0) doesn't have a "negative and positive solution", -0 = +0
So there is only one real solution
So in the top left graph, the discriminant is positive, since there are two distinct real solutions
The bottom left graph, the discriminant is negative, since the graph never crosses the x-axis, so the discriminant is negative, thus there are two imaginary solutions
The top right graph, the discriminant is zero, since there is only one real solution.
Answer:
a) positive
b) zero
c) negative
Step-by-step explanation:
Discriminant
[tex]b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0[/tex]
[tex]\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}[/tex]
[tex]\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}[/tex]
[tex]\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}[/tex]
Graph a
The parabola intersects the x-axis twice. Therefore, there are two real roots and so the discriminant is positive.
Graph b
The parabola intersects the x-axis once. Therefore, there is one real root and so the discriminant is zero.
Graph c
The parabola does not intersect the x-axis. Therefore, there are no real roots and so the discriminant is negative.
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