Lesson 10.5: Graphing Quadratic Equations in Two Variables

New Concept

We'll explore quadratic equations of the form $y = ax^2 + bx + c$. Their graphs are U-shaped curves called parabolas. By finding key features like the vertex and intercepts, you'll learn to sketch these graphs and find maximum or minimum values.

What’s next

Next, you’ll master finding a parabola's key features through interactive examples and practice cards, building up to graphing full equations and solving challenge problems.

Property

A quadratic equation in two variables, where $a, b,$ and $c$ are real numbers and $a \neq 0$, is an equation of the form

The graph of a quadratic equation is a U-shaped curve called a parabola.

Parabola Orientation
For the quadratic equation $y = ax^2 + bx + c$, if:

• $a > 0$, the parabola opens upward.
• $a < 0$, the parabola opens downward.

Examples

• The graph of $y = 2x^2 - 5x + 1$ opens upward because $a=2$, which is positive.

Property

For a parabola with equation $y = ax^2 + bx + c$:

• The axis of symmetry of a parabola is the line $x = -\frac{b}{2a}$.
• The vertex is on the axis of symmetry, so its $x$-coordinate is $-\frac{b}{2a}$.

To find the $y$-coordinate of the vertex, we substitute $x = -\frac{b}{2a}$ into the quadratic equation.

Examples

• For $y = x^2 - 6x + 5$, the axis of symmetry is $x = -\frac{-6}{2(1)} = 3$. The vertex's y-coordinate is $y = (3)^2 - 6(3) + 5 = -4$. The vertex is $(3, -4)$.

• For $y = -2x^2 + 8x - 1$, the axis of symmetry is $x = -\frac{8}{2(-2)} = 2$. The vertex's y-coordinate is $y = -2(2)^2 + 8(2) - 1 = 7$. The vertex is $(2, 7)$.

Property

To find the intercepts of a parabola with equation $y = ax^2 + bx + c$:

• y-intercept: Let $x=0$ and solve for $y$.
• x-intercepts: Let $y=0$ and solve for $x$.

The number of $x$-intercepts can be determined using the discriminant, $b^2 - 4ac$. If it's positive, there are two intercepts; if zero, there is one; if negative, there are none.

Examples

• For $y = x^2 - 5x + 4$, the y-intercept is $(0, 4)$. For the x-intercepts, set $y=0$: $0 = (x-4)(x-1)$, so the x-intercepts are $(4, 0)$ and $(1, 0)$.

• For $y = x^2 + 2x + 5$, the y-intercept is $(0, 5)$. The discriminant is $2^2 - 4(1)(5) = -16$. Since it's negative, there are no x-intercepts.

Property

How to graph a quadratic equation in two variables.

1. Determine if the parabola opens upward ($a > 0$) or downward ($a < 0$).
2. Find the axis of symmetry: $x = -\frac{b}{2a}$.
3. Find the vertex by substituting the axis of symmetry's $x$-value into the equation.
4. Find the $y$-intercept $(0, c)$ and its symmetric point across the axis of symmetry.
5. Find the $x$-intercepts by setting $y=0$ and solving for $x$.
6. Plot these key points and draw the parabola.

Examples

• For $y = x^2 - 2x - 3$: The parabola opens up ($a=1$). The axis of symmetry is $x = -\frac{-2}{2(1)} = 1$.

• For $y = -x^2 + 4x + 5$: The vertex x-coordinate is $x = -\frac{4}{2(-1)} = 2$. The y-coordinate is $y = -(2)^2 + 4(2) + 5 = 9$. The vertex is $(2, 9)$.

Property

The y-coordinate of the vertex of the graph of a quadratic equation is the

• minimum value of the quadratic equation if the parabola opens upward.
• maximum value of the quadratic equation if the parabola opens downward.

Examples

• For $y = 2x^2 + 8x + 3$, the parabola opens upward, so it has a minimum. The vertex is at $x = -\frac{8}{2(2)} = -2$. The minimum value is $y = 2(-2)^2 + 8(-2) + 3 = -5$.

• For $y = -x^2 + 6x - 4$, the parabola opens downward, so it has a maximum. The vertex is at $x = -\frac{6}{2(-1)} = 3$. The maximum value is $y = -(3)^2 + 6(3) - 4 = 5$.