The equation is
.
We are looking for a function with a vertex above the x-axis and a function that opens upward (has coefficient a > 0).
The first function opens downward and intersects the x-axis. The second function has a vertex below the x-axis. The third function satisfies our requirements. The fourth function has a vertex on the x-axis.
We can solve this algebraically with the knowledge that the real solutions of a quadratic are its x-intercepts. If there are no x-intercepts (because it lies entirely above or below the x-axis), then there are no real solutions. This is true when the discriminant
. You can see that from the quadratic formula. This holds true for both answers A and C, so to find the correct one, we remember that when the coefficient a of the
term is positive, the graph opens upwards, so we choose
C.
One day 6 said, "I wish I were an adult." And 1/3 came along and said, "I can do a magic flip! If I do this special flip and you're brave enough to combine our powers together we'll be 18!" 6 thought this was a brilliant idea so 1/3 flipped on his head and turned into a 3 and when 6
I think the answer is 6 I’m not sure tho
Answer:
x = -5/2
Step-by-step explanation:
From the quadratic equation
the axis of symmetry for quadratics of the form
y = ax² + bx + c is
x = -b/2a
x² + 5x + 9
x = -5/2
Answer:
D
Step-by-step explanation: