Question

\[f(x)=x^2-10x-56\] 1) What are the zeros of the function? Write the smaller \[x\] first, and the larger \[x\] second. \[\text{smaller }x=\] \[\text{larger }x=\] 2) What is the vertex of the parabola?

Answer 1

The zeros of the function f(x) = x^2 - 10x - 56 are -4 and 14. The vertex of the parabola is at the point (5, -36).

Step-by-step explanation:

To find the zeros of the quadratic function f(x) = x^2 - 10x - 56, we can factor the equation or use the quadratic formula. The factored form of the equation is (x - 14)(x + 4) = 0.

Setting each factor equal to zero gives us the zeros of the function. Therefore:

• smaller x = -4
• larger x = 14

To find the vertex of the parabola, we use the vertex formula for a quadratic function in the form y = ax^2 + bx + c, which is (-b/2a, f(-b/2a)).

For our function, a = 1 and b = -10.

Plugging these into the formula, the x-coordinate of the vertex is 10/2(1) or 5.

To find the corresponding y-coordinate, we substitute x with 5 back into the function to get f(5) = 5^2 - 10(5) - 56 which simplifies to -36. Therefore, the vertex of the parabola is (5, -36).