Question

In the xy-plane, a parabola has vertex (3, 1) and
intersects the x-axis at two points. If the equation of
the parabola is written in the form y = -ax² +
bx+c, where a, b, and c are constants, which of the
following could be a value of c?
A) -8
B) 2
C) 3
D) 7

Answer 1

We will see that the parabola must open downwards, then the value of c must be negative, and thus, the correct option is A, c = -8.

Which could be the value of c?

The general quadratic equation is written as:

y = a*x^2 + b*x + c

Where a is the leading coefficient, and it defines if the parabola opens up or down, depending on its sign.

And we know that we have two real solutions, so the discriminant (4*a*c in the general equation) must be positive.

Then:

4*a*c> 0

Now, notice that the vertex of the parabola is above the x-axis, then the parabola only will intercept the x-axis if it opens downwards, then a must be negative.

Then the value of c also must be negative, due to the above inequality.

Then the only option that could be the value of c is option A, c = -8.

Learn more about quadratic equations:

https://brainly.com/question/1214333

#SPJ1