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Quadratic function:

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 ≠ 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 = − 𝑏 2 𝑎 .
Roots (or Zeros): The points where the graph intersects the x-axis. These are the solutions to the equation 𝑎 𝑥 2 + 𝑏 𝑥 + 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 2 − 4 𝑥 + 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/s 2 ).
32 𝑡 corresponds to the upward initial velocity of the ball ( 32 ft/s ).
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 = − 𝑏/ 2 𝑎 . 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 − 16 𝑡 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.