Quadratic function, or the vertex form of a quadratic function. The standard form of a quadratic function (a parabola) is:

y = ax² + bx + c

The vertex form of a quadratic function (a parabola) is:

y = a(x – h)² + k

The quadratic function for a parabola in standard form can be converted to vertex form by completing the square.



I am going to use the standard form of a quadratic function:

y = ax² + bx + c

The standard form of quadratic equation would be:

ax² + bx + c = 0

Solve for x using the quadratic formula:

x = (-b/2a) ± [√(b² - 4ac)/2a]

the axis of symmetry (the x coordinate of the vertex) of the parabola is:

x_axis_of_symmetry = (-b/2a)

Since the axis of symmetry must be x = 1:

(-b/2a) = x_axis_of_symmetry

(-b/2a) = 1

Choose any value for a, and then solve for b

Or

Choose any value for b, and then solve for a

As an example, since the parabola must open downward, a must be negative.

let a = -1

(a = -1 is an arbitrary choice. You can choose any value for “a”, as long as it is a negative value.)

(-b/2a) = 1

-b/[2*(-1)] = 1

-b/(-2) = 1

b/(2) = 1

[b/(2)]*(2) = 1*(2)

b*(2/2) = 1*2

b*(1) = 1*2

b = 2

y = ax² + bx + c

substituting the values a = -1 and b = 2

y = -1*x² + 2*x + c

y = -x² + 2x + c

you can now choose a value for c

Strictly speaking, you can choose any value for “c”. However, your problem statement asks that you compute the x-intercepts. “c” must be a positive value for the quadratic function to cross the x-axis. So I recommend you choose a positive value for “c”.

for example, let c = 5

using the values a = -1, b = 2, and c = 5

>>> The standard form of the final quadratic function is:

y = -x² + 2x + 5

>>> The standard form of the final quadratic equation is:

0 = -x² + 2x + 5

>>> The vertex form of the final quadratic function is:

y = -(x – 1)² + 6

>>> The vertex form of the final quadratic equation is:

0 = -(x – 1)² + 6

kelsey your question​