Question

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Answer 1

The equation is $y=(x-7)^2+7$.

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 $b^2-4ac \ \textless \ 0$. 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 $x^2$ term is positive, the graph opens upwards, so we choose C.

Answer 2

Answer:
choice c is the correct one

Explanation:
The equation that lies completely above the x-axis should be the one having its y-values greater than zero.

squaring a number will certainly give a positive value.
looking at the choices:
choice a: might give a -ve value if the squaring gives a number greater than 7
choice b: might give a -ve value if the squaring gives a number less than 7
choice c: will always be positive
choice d: might give a zero at x=7