Graphing and Intercepts

Main idea

The graph of an equation is a picture of every point $(x, y)$ that makes it true. The first points worth finding are where it crosses the axes, and each one comes from setting the other variable to zero.

\text{x-intercept: } ({\htmlData{sym=0}{a}}, 0), \qquad \text{y-intercept: } (0, {\htmlData{sym=1}{b}})

What the symbols mean

$a$
x-intercept
A place the graph crosses the x-axis; y is 0 there.
$b$
y-intercept
Where the graph crosses the y-axis; x is 0 there.

Explanation

A graph is simply a picture of all the points that make an equation true, so you can see at a glance what the equation does. An intercept is a spot where that picture crosses one of the axes. The formula up top has two pieces: $(a, 0)$ (an x-intercept, where the height is zero) and $(0, b)$ (the y-intercept, where you haven't moved left or right at all). In real problems these are often the answers people actually want, like where a ball lands or what a quantity starts at. That's the whole idea.

The coordinate plane. A point $(x, y)$ sits $x$ units right of the origin and $y$ units up (negative means left or down). The axes split the plane into four quadrants, numbered counterclockwise starting from the upper right.

Graphing by plotting points. Make a table of $x$ values, work out each $y$ , plot them and connect them smoothly. It's slow, but it always works, and it's how you check any shortcut you learn later.

Intercepts. Think about what's true on each axis. Every point on the x-axis has height $0$ , and every point on the y-axis has $x = 0$ . So:

x-intercepts: set $y = 0$ and solve for $x$ . There can be none, one or many.
y-intercept: set $x = 0$ . A function has at most one, since one input can only give one output.

Is a point on the graph? A point lies on a graph exactly when its coordinates make the equation true. So "is $(2, -1)$ on the graph?" is answered by plugging in and checking.

See it

This is a ball's path, y = −0.05x² + x + 2, in meters. Notice the intercepts are the interesting points: x = 0 gives the y-intercept, the launch height of 2 m, and y = 0 gives the x-intercept, where the ball lands at x = 10 + √140 ≈ 21.8 m. In between it peaks at 7 m, 10 m out.

The parabola y = x² − 4x − 5 crosses the x-axis at x = −1 and x = 5, where y = 0, and the y-axis at y = −5, where x = 0.

Worked examples

Example 1

Find the intercepts of $y = x^2 - 4x - 5$ .

1
For the x-intercepts set $y = 0$ and factor: $(x - 5)(x + 1) = 0$ , so $x = 5$ or $x = -1$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Which quadrant contains $(-3, 5)$ ?

Problem 2+5 XP

Find the x-intercepts of $y = x^2 - 4x - 5$ .

Problem 3+5 XP

Find the y-intercept of $y = x^2 - 4x - 5$ .

Problem 4+5 XP

Find the intercepts of $3x + 4y = 12$ . Enter the x-intercept, then the y-intercept.

Problem 5+5 XP

Is $(2, -1)$ on the graph of $y = x^2 - 5$ ?

Problem 6+5 XP

A ball's height is $h = -5t^2 + 20t + 25$ meters after $t$ seconds. When does it hit the ground?

Problem 7+5 XP

What is the area of the triangle that the line $3x + 4y = 12$ makes with the two axes?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A parabola $y = ax^2 + bx + c$ has x-intercepts 2 and 4, and its y-intercept is $-8$ . Find $a$ , $b$ and $c$ .

Where it's used

Intercepts answer real questions: when a thrown ball lands ( $h = 0$ ), the break-even point where profit is zero, and the starting value of a quantity at time 0.

Projectile MotionAlgebra · Physics

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