Another music store rents trombone for $30 per month plus a yearly fee of $48. Which deal is better? Should Maria change her ren
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So y=2x is the only answer that fit y=kx, while k=2
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.
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