Graphing Linear & Quadratic Functions: Intercepts, Maxima & Minima - Algebra 1

Section 1

Where Did Graphs Come From?

People have been solving equations for thousands of years, but for most of history they did it entirely with words and numbers — no pictures at all. The idea of plotting an equation on a grid so you could actually see its shape was a breakthrough that changed mathematics forever. Here's how it happened.

~300 BCE • Ancient Greece

Euclid and other Greek mathematicians studied lines, circles, and parabolas using pure geometry. They understood curves, but they had no coordinate system to graph equations on.

1637 • René Descartes

French philosopher and mathematician René Descartes published La Géométrie, introducing the Cartesian coordinate plane — the x-y grid you use today. For the first time, algebra and geometry were connected: every equation could become a picture, and every picture could become an equation.

1700s • Euler & the Function Concept

Leonhard Euler formalized the idea of a function — a rule that takes an input and produces exactly one output. He used the familiar f(x) notation that you still see in every algebra class.

1800s–1900s • Classroom Standard

As public education expanded, graphing linear and quadratic functions became a core skill. Textbooks adopted the standard approach of making tables, plotting points, and connecting them to reveal a line or parabola.

The question this lesson answers is straightforward but powerful: given a linear or quadratic equation, how do you turn it into a graph, and once you have the graph, how do you identify its intercepts, maximum, or minimum?

Section 2

Core Definitions & Principles

Before we start graphing, you need a handful of key vocabulary words. Each one describes a specific feature of a function's graph that you'll be asked to identify over and over in algebra.

Linear Function

A function whose graph is a straight line. Its general form is f(x) = mx + b, where m is the slope and b is the y-intercept. The line goes on forever in both directions.

Quadratic Function

A function whose graph is a U-shaped curve called a parabola. Its standard form is f(x) = ax² + bx + c. If a > 0 the parabola opens upward; if a < 0 it opens downward.

Intercepts

Points where the graph crosses an axis. The y-intercept is where the graph crosses the y-axis (set x = 0). The x-intercept(s) are where the graph crosses the x-axis (set y = 0).

Maximum & Minimum

The highest or lowest point on a parabola is called the vertex. If the parabola opens downward, the vertex is a maximum. If it opens upward, the vertex is a minimum. Linear functions have neither — they keep going up or down forever.

Axis of Symmetry

A vertical line that divides a parabola into two mirror-image halves. Its equation is x = −b / (2a). The vertex always sits on this line.

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Key Takeaway

Think of a function's graph like a road on a map. Intercepts are the spots where that road crosses a major highway (the axes). A maximum or minimum is the highest hilltop or the deepest valley along the road. Being able to read these features tells you the most important information about any function at a glance — just like knowing where the intersections and hills are tells a driver what's ahead.

Section 3

Visual Guide: Lines & Parabolas on the Coordinate Plane

The diagram below shows a linear function and a quadratic function graphed on the same coordinate plane, with every key feature labeled. Study it carefully — you'll use these same labels whenever you analyze a function.

Figure 1 — A linear function (cyan line) and a quadratic function (violet parabola) on the same coordinate plane, with intercepts, vertex, and axis of symmetry labeled.

Notice a few things in the diagram above. The linear function y = x + 1 is a perfectly straight line. It has exactly one x-intercept and one y-intercept. It never turns around, so it has no maximum or minimum. The quadratic function y = −x² + 4x − 1 curves into an upside-down U shape because the leading coefficient (the number in front of x²) is negative. Its highest point — the vertex at (2, 3) — is the maximum. The dashed vertical line through the vertex is the axis of symmetry, and the left half of the parabola is a mirror image of the right half.

Section 4

The Equations You Need

Let's look at the formulas that let you find intercepts, slope, and vertex without relying on a graph alone.

Slope-Intercept Form (Linear)

y = mx + b

m = slope (rise ÷ run) | b = y-intercept

In the equation above, b gives you the y-intercept directly — just read it off. To find the x-intercept, set y = 0 and solve for x: the x-intercept is at x = −b / m. The slope m tells you how steeply the line rises or falls. A positive slope means the line goes up from left to right; a negative slope means it goes down.

Standard Form (Quadratic)

y = ax² + bx + c

a = direction & width | b, c = coefficients | y-intercept = c

For a quadratic, the y-intercept is simply the constant c. That's because when x = 0, the equation becomes y = a(0)² + b(0) + c = c. Finding x-intercepts requires more work — you either factor, complete the square, or use the quadratic formula.

Vertex Formula

x = −b / (2a)

Gives the x-coordinate of the vertex. Plug it back into the original equation to get the y-coordinate.

The vertex formula is one of the most useful tools in Algebra 1. Once you know x = −b / (2a), you substitute that value into y = ax² + bx + c to find the y-coordinate of the vertex. If a > 0 the vertex is a minimum; if a < 0 the vertex is a maximum.

Quadratic Formula (for x-intercepts)

x = (−b ± √(b² − 4ac)) / (2a)

The discriminant b² − 4ac determines how many x-intercepts exist: positive → 2, zero → 1, negative → 0.

The expression under the square root, b² − 4ac, is called the discriminant. It acts as a preview: if the discriminant is positive, the parabola crosses the x-axis twice. If it equals zero, the vertex itself sits on the x-axis (one x-intercept). If the discriminant is negative, the parabola never touches the x-axis at all.

Section 5

Deep Dive: Comparing Features of Linear vs. Quadratic Graphs

The table below summarizes every feature you should check whenever you analyze a function's graph. Use it as a reference when you work through problems.

Feature	Linear (y = mx + b)	Quadratic (y = ax² + bx + c)
Shape	Straight line	Parabola (U or upside-down U)
y-intercept	Always exactly 1, at (0, b)	Always exactly 1, at (0, c)
x-intercept(s)	Exactly 1 (unless slope = 0 and b ≠ 0)	0, 1, or 2 depending on the discriminant
Vertex	None — line extends forever	Exactly 1, at (−b/(2a), f(−b/(2a)))
Max or Min	Neither — range is all real numbers	Max if a < 0; Min if a > 0
Axis of Symmetry	None	x = −b / (2a)
End Behavior	Both ends go to ±∞ in opposite directions	Both ends go to +∞ (opens up) or −∞ (opens down)

Figure 2 — Three parabola scenarios based on the discriminant: two x-intercepts, one x-intercept, and no x-intercepts.

Figure 2 illustrates the three cases you can encounter with a quadratic function. In the left panel the parabola opens upward and dips below the x-axis, producing two x-intercepts and a minimum. The center panel shows a downward-opening parabola whose vertex just touches the x-axis — this gives exactly one x-intercept and a maximum. In the right panel the parabola never reaches the x-axis at all, so there are zero x-intercepts, and the vertex is still a minimum that's positioned above the x-axis.

Discriminant Spectrum: Number of x-Intercepts

b² − 4ac < 0 → 0 intercepts b² − 4ac > 0 → 2 intercepts

Section 6

Worked Example

Let's walk through a complete problem from start to finish. We'll graph the quadratic function f(x) = 2x² − 8x + 6 and identify all key features.

Graphing f(x) = 2x² − 8x + 6

1
Step 1 — Identify the y-intercept

Set x = 0 and evaluate:
f(0) = 2(0)² − 8(0) + 6 = 6

The y-intercept is (0, 6).
2
Step 2 — Find the vertex

Use x = −b / (2a). Here a = 2 and b = −8:
x = −(−8) / (2 × 2) = 8 / 4 = 2
Now plug x = 2 back in to find the y-coordinate:
f(2) = 2(2)² − 8(2) + 6 = 8 − 16 + 6 = −2

The vertex is (2, −2). Since a = 2 > 0, the parabola opens upward, so this vertex is a minimum.
3
Step 3 — Find the x-intercepts

Set f(x) = 0 and solve 2x² − 8x + 6 = 0. First, divide everything by 2:
x² − 4x + 3 = 0
Factor:
(x − 1)(x − 3) = 0

So x = 1 or x = 3. The x-intercepts are (1, 0) and (3, 0).
4
Step 4 — Axis of symmetry

The axis of symmetry passes through the vertex.

x = 2. Notice that the two x-intercepts (x = 1 and x = 3) are equidistant from x = 2 — that's the symmetry at work.
5
Step 5 — Plot and sketch

Plot the five key points: the y-intercept (0, 6), the two x-intercepts (1, 0) and (3, 0), the vertex (2, −2), and its mirror of the y-intercept at (4, 6). Connect them with a smooth U-shaped curve. Draw the dashed axis of symmetry at x = 2.
6
Final Answer

y-intercept: (0, 6) | x-intercepts: (1, 0) and (3, 0) | Minimum: (2, −2) | Axis of symmetry: x = 2

Section 7

Strengths & Limitations

Graphing is one of the most powerful tools in algebra, but it's worth knowing where each method shines and where it falls short.

Method	Strengths	Limitations
Graphing by hand	Builds deep understanding of shape and features; great for seeing the big picture	Slow; accuracy limited by your drawing; hard with decimals or large numbers
Using the vertex formula	Gives the exact vertex quickly; works for every quadratic	Doesn't directly give x-intercepts; still need separate step
Factoring	Fast when the quadratic factors neatly; gives exact x-intercepts	Many quadratics don't factor over integers
Quadratic formula	Works for every quadratic; reveals discriminant info	More arithmetic to manage; easier to make sign errors
Graphing calculator / software	Fast, accurate, great for checking work	Doesn't build algebraic skills; answers may be approximate

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Key Takeaway

Think of these methods like different tools in a toolbox. A screwdriver and a power drill can both drive screws, but each one works better in different situations. In algebra, graphing gives you the visual overview, formulas give you precise numbers, and factoring gives you speed — when it works. The best students learn to pick the right tool for each problem rather than relying on just one approach.

Section 8

Where This Leads Next

The skills you're building right now — finding intercepts, identifying maxima and minima, and understanding how an equation's coefficients control a graph's shape — are the foundation for nearly everything that comes next in math. Here's a preview of how this lesson connects to more advanced topics.

What You Learn Now	Where It Goes Next
Finding the vertex of a parabola	In Algebra 2, you'll rewrite quadratics in vertex form y = a(x − h)² + k and use completing the square
Graphing y = mx + b	In later courses, you'll graph systems of two or more linear equations and find their intersection points
Using the discriminant	In Pre-Calculus, the discriminant helps you analyze complex (imaginary) roots when b² − 4ac < 0
Identifying max/min by sign of a	In Calculus, you'll use derivatives to find maxima and minima of any function, not just quadratics
Plotting points from a table	In statistics, you'll create scatter plots and fit regression lines or curves to real data

The point is that linear and quadratic functions aren't just isolated topics — they are the simplest members of a huge family of functions you'll study over the next several years. Mastering them now means you'll have a much easier time when the functions get more complicated. The ideas of intercepts, vertex, and end behavior carry over directly to polynomials, exponentials, and beyond.

Section 9

Practice Problems

Try each problem on paper first, then click "Show Answer" to check your work. The problems get progressively harder.

Question 1: Problem 1 — Conceptual

A quadratic function has a = −3. Without graphing or calculating anything else, does the function have a maximum or a minimum? How do you know?

Show answer

The function has a maximum. When the leading coefficient a is negative, the parabola opens downward — like an upside-down U. The very top of that curve is the vertex, and since the graph falls away on both sides, the vertex is the highest point the function can reach. That highest point is the maximum.

Question 2: Problem 2 — Basic Identification

Find the slope, y-intercept, and x-intercept of the linear function y = −3x + 9.

Show answer

Slope: m = −3 (the line falls 3 units for every 1 unit to the right). y-intercept: Set x = 0 → y = −3(0) + 9 = 9. The y-intercept is (0, 9). x-intercept: Set y = 0 → 0 = −3x + 9 → 3x = 9 → x = 3. The x-intercept is (3, 0).

Question 3: Problem 3 — Intermediate

For the quadratic function f(x) = x² + 6x + 5, find the vertex, state whether it is a maximum or minimum, find the y-intercept, and find the x-intercepts.

Show answer

Vertex: x = −b / (2a) = −6 / (2 × 1) = −3. Then f(−3) = (−3)² + 6(−3) + 5 = 9 − 18 + 5 = −4. Vertex: (−3, −4). Max or Min: Since a = 1 > 0, the parabola opens upward, so the vertex is a minimum. y-intercept: f(0) = 0 + 0 + 5 = 5 → (0, 5). x-intercepts: Set x² + 6x + 5 = 0. Factor: (x + 1)(x + 5) = 0. So x = −1 or x = −5 → (−1, 0) and (−5, 0).

Question 4: Problem 4 — Applied / Multi-Step

A ball is thrown upward from a 4-foot platform. Its height (in feet) after t seconds is modeled by h(t) = −16t² + 32t + 4. Find the maximum height the ball reaches and the time(s) when it hits the ground (h = 0). Round to two decimal places if needed.

Show answer

Maximum height: t = −b / (2a) = −32 / (2 × −16) = −32 / −32 = 1 second. h(1) = −16(1)² + 32(1) + 4 = −16 + 32 + 4 = 20 feet. The maximum height is 20 feet, reached at t = 1 second. Hits the ground: Set −16t² + 32t + 4 = 0. Using the quadratic formula: t = (−32 ± √(32² − 4(−16)(4))) / (2(−16)) = (−32 ± √(1024 + 256)) / (−32) = (−32 ± √1280) / (−32). √1280 ≈ 35.78. So t = (−32 + 35.78) / (−32) ≈ −0.12 (reject — negative time) or t = (−32 − 35.78) / (−32) ≈ 2.12 seconds. The ball hits the ground at approximately t ≈ 2.12 seconds.

Question 5: Problem 5 — Critical Thinking

A student says, "If a quadratic function has no x-intercepts, then it has no maximum or minimum either." Is this statement true or false? Explain your reasoning, and give an example to support your answer.

Show answer

The statement is false. Every quadratic function has a vertex, and that vertex is always either a maximum or a minimum — regardless of whether the graph crosses the x-axis. The number of x-intercepts depends on the discriminant, but the vertex exists for every parabola. Example: Consider f(x) = x² + 2. Here a = 1 > 0, so the parabola opens upward. The vertex is at x = −0/(2×1) = 0, and f(0) = 2. The vertex (0, 2) is a minimum. However, setting x² + 2 = 0 gives x² = −2, which has no real solutions — there are no x-intercepts. So the function has a minimum but no x-intercepts, disproving the student's claim.

Summary

Lesson Summary

Summary

A linear function y = mx + b produces a straight-line graph. Its y-intercept is at (0, b), its x-intercept is at (−b/m, 0), and its slope m controls the steepness and direction. Linear functions have no maximum or minimum because the line extends infinitely in both directions.

A quadratic function y = ax² + bx + c produces a curved graph called a parabola. Its y-intercept is at (0, c). It can have 0, 1, or 2 x-intercepts, determined by the discriminant b² − 4ac. The most important point on the parabola is the vertex, found at x = −b/(2a). If a > 0 the parabola opens upward and the vertex is a minimum; if a < 0 it opens downward and the vertex is a maximum. The vertical line through the vertex, x = −b/(2a), is the axis of symmetry. Together, these features — intercepts, vertex, direction, and symmetry — give you a complete picture of any linear or quadratic function.