Lesson 96: Graphing Quadratic Functions

New Concept

The equation of the axis of symmetry and the $x$-coordinate of the vertex of a quadratic function is $x = -\frac{b}{2a}$.

What’s next

Next, you’ll use this formula to find the vertex and axis of symmetry, the key points for graphing any parabola.

Ever wonder how to find the exact moment a thrown object reaches its peak? This example demonstrates how the vertex formula answers that question directly.

Example Problem

A flare is launched from a boat with an initial velocity of 80 feet per second. The flare is launched from a height of 6 feet above the water. At what time does the flare reach its maximum height?

Property

The equation of the axis of symmetry for a quadratic function in the form $f(x) = ax^2 + bx + c$ is $x = -\frac{b}{2a}$. To find the vertex, substitute this $x$-value back into the function.

Explanation

Think of the axis of symmetry as the parabola's spine! This magic formula pinpoints the center line, leading you directly to the vertex—the function's peak or valley. It's the ultimate shortcut to sketching a perfect parabola without plotting a million points. Find the center, find the vertex, and you're halfway to a flawless graph!

Examples

Find the vertex of $y = x^2 + 6x + 8$. Axis of symmetry: $x = -\frac{6}{2(1)} = -3$. Vertex: $y = (-3)^2 + 6(-3) + 8 = -1$. The vertex is $(-3, -1)$.
Find the vertex of $y = 2x^2 + 12x + 10$. Axis of symmetry: $x = -\frac{12}{2(2)} = -3$. Vertex: $y = 2(-3)^2 + 12(-3) + 10 = -8$. The vertex is $(-3, -8)$.
Find the vertex of $y = 4x^2 + 8$. Since $b=0$, the axis of symmetry is $x = -\frac{0}{2(4)} = 0$. The vertex is $(0, 8)$.

Let's transform this quadratic equation into a complete graph by finding its essential features. This example showcases how to graph a function in the form $y = ax^2 + bx + c$.

Example Problem

Graph the function $y = 2x^2 + 8x + 5$.

Property

A zero of a function is an $x$-value for a function where $f(x) = 0$. It is the point where the graph of the function meets or intersects the $x$-axis. It's another name for an $x$-intercept.

Explanation

Zeros are just a cool name for the $x$-intercepts! They're the spots where your parabola either touches or crosses the horizontal x-axis. Finding them means you're solving the puzzle of where the function's output hits zero. A parabola can have two zeros, one zero, or even none if it's too shy to meet the x-axis!

Examples

The graph of $y = x^2 + x - 6$ crosses the x-axis at $x = -3$ and $x = 2$. The zeros are $-3$ and $2$.
The graph of $y = x^2 - 8x + 16$ touches the x-axis at a single point, $x = 4$. The only zero is $4$.
The graph of $y = -x^2 - 5$ floats entirely below the x-axis and never crosses it. This function has no real zeros.

Property

The height $h$ in feet of an object after $t$ seconds is given by the formula $h = -16t^2 + vt + s$, where $v$ is the initial vertical velocity in feet per second and $s$ is the starting height in feet.

Explanation

Ever wonder how high a ball will fly? This formula is your crystal ball for physics! It models the classic up-and-down arc of a thrown object. The $-16t^2$ part is gravity doing its thing, pulling the object back down. Just plug in the starting speed and height, and you can predict its entire flight path!

Examples

A ball is thrown up at 40 feet per second from 5 feet high. Its flight path is modeled by the function $h = -16t^2 + 40t + 5$.
To find the maximum height for $h = -16t^2 + 40t + 5$, find the time at the vertex: $t = -\frac{40}{2(-16)} = 1.25$ seconds.
A soccer ball is kicked with a path modeled by $f(x) = -8x^2 + 16x$. The time to reach maximum height is $x = -\frac{16}{2(-8)} = 1$ second.