A circle has a sector with area \dfrac{1}{2}\pi 2 1 πstart fraction, 1, divided by, 2, end fraction, pi and central angle of \
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<h3>Answer:</h3>
The Slope is
Notice that both the points, and
, are on the line. So we can use those points to calculate the slope. Recall that that slope of the line
can be calculated by
if we have the points,
and
.
Given:
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