In the xy𝑥𝑦-plane, a parabola has vertex (−25, 78)(-25, 78) and intersects the x𝑥-axis at two points. If the equation of the parabola is written in the form y=ax2+bx+c𝑦=𝑎𝑥2+𝑏𝑥+𝑐, where a𝑎, b𝑏, and c𝑐 are constants, which of the following could be the value of a+b+c𝑎+𝑏+𝑐?−80-80eliminate−78-78eliminate00eliminate18

The vertex form of a parabola is given by y = a(x - h)² + k, where (h, k) is the vertex of the parabola. Given that the vertex is (-25, 78), we can substitute these values into the vertex form to get y = a(x + 25)² + 78.

The parabola also intersects the x-axis at two points. This means that the y-value at these points is 0. So, we can set y = 0 in our equation to solve for a:

0 = a(x + 25)² + 78 -78 = a(x + 25)² a = -78 / (x + 25)²

Since we don't know the x-values where the parabola intersects the x-axis, we can't find a specific value for a. However, we know that a must be negative because the parabola opens downwards (since it intersects the x-axis at two points and has a maximum at y = 78).

Now, let's expand the vertex form of the parabola to get it in the form y = ax² + bx + c:

y = a(x² + 50x + 625) + 78 y = ax² + 50ax + 625a + 78

Comparing this with y = ax² + bx + c, we can see that b = 50a and c = 625a + 78.

The question asks for the value of a + b + c. Substituting b and c, we get:

a + b + c = a + 50a + 625a + 78 = 676a + 78

Since a is negative, a + b + c is also negative. Therefore, the only possible answer is -80.