What is y in y = x^2 + x?

In mathematics, the equation $y = x^2 + x$ represents a quadratic function. This type of function is a polynomial of degree 2, which means it has an $x^2$ term. Let’s break down what this equation means and how we can understand it better.

Understanding the Equation

Components of the Equation

  • $y$: This is the dependent variable, which means its value depends on the value of $x$
  • $x$: This is the independent variable, which you can choose freely.
  • $x^2$: This term means $x$ is squared, or multiplied by itself.
  • $x$: This term is added to $x^2$

General Form of a Quadratic Function

A quadratic function generally has the form $y = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants. In our equation, $a = 1$, $b = 1$, and $c = 0$. This means:

  • The coefficient of $x^2$ is 1.
  • The coefficient of $x$ is 1.
  • There is no constant term (or you could say the constant term is 0).

Graphing the Equation

When you graph $y = x^2 + x$, you get a parabola. A parabola is a U-shaped curve that can open either upwards or downwards. In this case, since the coefficient of $x^2$ is positive (1), the parabola opens upwards.

Key Features of the Graph

  • Vertex: The highest or lowest point of the parabola. For $y = x^2 + x$, the vertex can be found using the formula $x = -frac{b}{2a}$. Plugging in $a = 1$ and $b = 1$, we get $x = -frac{1}{2 times 1} = -frac{1}{2}$. To find the y-coordinate of the vertex, substitute $x = -frac{1}{2}$ back into the equation: $y = bigg(-frac{1}{2}bigg)^2 + bigg(-frac{1}{2}bigg) = frac{1}{4} – frac{1}{2} = -frac{1}{4}$. So, the vertex is at $(-frac{1}{2}, -frac{1}{4})$
  • Axis of Symmetry: The vertical line that passes through the vertex, which in this case is $x = -frac{1}{2}$
  • Y-intercept: The point where the graph crosses the y-axis. This occurs when $x = 0$. Plugging $x = 0$ into the equation, we get $y = 0^2 + 0 = 0$. So, the y-intercept is at $(0, 0)$
  • X-intercepts: The points where the graph crosses the x-axis. These can be found by solving the equation $x^2 + x = 0$. Factoring, we get $x(x + 1) = 0$. So, $x = 0$ or $x = -1$. Thus, the x-intercepts are at $(0, 0)$ and $(-1, 0)$

Real-World Applications

Quadratic functions like $y = x^2 + x$ appear in various real-world scenarios. For example, they can describe the trajectory of a ball thrown into the air, the shape of satellite dishes, and the design of certain types of bridges.

Conclusion

Understanding the equation $y = x^2 + x$ involves recognizing it as a quadratic function and being able to graph it to reveal its key features. These features include the vertex, axis of symmetry, y-intercept, and x-intercepts. Recognizing the practical applications of quadratic functions can help you appreciate their importance in both mathematics and the real world.

4. Wikipedia – Quadratic Function

Citations

  1. 1. Khan Academy – Quadratic Functions
  2. 2. Math is Fun – Quadratic Equations
  3. 3. Purplemath – Quadratic Functions

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ