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A quadratic function is a type of mathematical function that can be expressed in the standard form: $$ f(x) = ax^2 + bx + c $$ Here,

a, b, and c are constants (with $a \neq 0$),
x is the variable,
and the term $a x^2 $ represents the quadratic term, which gives the function its characteristic parabolic shape.

The graph of a quadratic function is called a parabola. If 𝑎 > 0 , the parabola opens upwards, and if 𝑎< 0, it opens downwards.

Components of a Quadratic Function:

Vertex: The highest or lowest point on the graph, depending on the direction the parabola opens.
Axis of Symmetry: A vertical line that passes through the vertex and divides the parabola into two symmetric halves. Its equation is \[x = - \frac{- b}{2a}\]
Roots (or Zeros): The points where the graph intersects the x-axis. These are the solutions to the equation \[ax^2 + bx +c = 0\]
Y-Intercept: The point where the graph intersects the y-axis, which is at (0, 𝑐).

Examples:

Example 1:

Let $$ f(x) = x^2 - 4x + 3 $$

Here, 𝑎 = 1, 𝑏 = − 4, and 𝑐 = 3.
The graph is a parabola opening upwards ( 𝑎 > 0 ).
The axis of symmetry is $$ x = - \frac{- 4}{2(1)} = 2 $$
The roots can be found by factoring: \[x^ - 4x + 3 = = (x - 3)(x - 1)\] So, the roots are 𝑥 = 1 and 𝑥 = 3.

Example 2:

Let $$ f(x) = - 2x^ 2 + 8x - 5 $$

Here, 𝑎 = − 2, 𝑏 = 8, and 𝑐 = −5.
The graph is a parabola opening downwards (𝑎<0).
The axis of symmetry is $$ x = - \frac{8}{2(-2)} = 2 $$
The vertex can be found by substituting 𝑥 = 2 into the function: $$ f(2) = - 2 (2^2) + 8(2) - 5 = 3 $$ So, the vertex is at $(2, 3)$.

Example 3
Real life example: Quadratic functions are often used to model projectile motion. For instance, if a ball is thrown upwards, its height ℎ(𝑡) at time 𝑡 can be described by a quadratic function like $$ h(t) = - 16 t^2 + 32 t + 5. $$ Where,

$- 16 t^2$accounts for the downward acceleration due to gravity ( $- 16 ft/sec^2$).
$ 32 t$ corresponds to the upward initial velocity of the ball ( $32 ft/sec$).
5 is the initial height of the ball at $t = 0$.

This parabola opens downward because the coefficient of $t^2$is negative (−16), meaning the ball reaches a peak height before falling back to the ground.

Features of Projectile Motion Modeled by Quadratics:

Vertex: Represents the peak height of the ball and occurs at time $t = \frac{- b}{2a}$. Plugging in the values: $$ t = - \frac{32}{2(-16)} = 1 \quad second $$ Substituting $t = 1$ into the equation: $$ h(1) = - 16 (1^2) + 32(1) + 5 = 21 \quad feet $$ The ball reaches a maximum height of 21 feet at $t = 1$ second.
Roots: Where $h(t) = 0$, the ball hits the ground. Solving the equation $$ - 16t^2 + 32 t + 5 = 0 $$ gives the times at which this happens.
Trajectory: The graph of the quadratic function visually shows the complete motion of the ball—up to its peak and back down.