Quadratic Functions 1 - Vertex Form PDF

# Degree: 2 Quadratic Functions ## Standard Form: $y = ax^2 + bx + c$ - Graph: Parabola - Opens **upward** if $a > 0$ - Opens **downward** if $a < 0$ - **a** determines the opening ## Is $y = (x - 4)(x + 3)$ a quadratic function? Yes! - $y = x^2 + 3x - 4x -12$ - $y = x^2 - x - 12$ - If **a** is...

# Degree: 2 Quadratic Functions ## Standard Form: $y = ax^2 + bx + c$ - Graph: Parabola - Opens **upward** if $a > 0$ - Opens **downward** if $a < 0$ - **a** determines the opening ## Is $y = (x - 4)(x + 3)$ a quadratic function? Yes! - $y = x^2 + 3x - 4x -12$ - $y = x^2 - x - 12$ - If **a** is positive, the parabola opens up ## Parts of a Parabola - **Vertex (p, q):** the lowest point (minimum) or highest point (maximum) - **y-intercept** - **x-intercept** - **Line of symmetry** ## Vertex Form: $y = ±(x - p)^2 + q$ - **±** determines the opening - **(p, q)** is the vertex ## $y = (x - 4)^2 + 3$ (Example 1) - Opening: **upward** - Vertex: **(4, 3)** - No **x-intercept** - **y-intercept**: let $x = 0$ * $y = (0 - 4)^2 + 3$ * $y = 19$ * Therefore, the **y-intercept** is 19. - Line of symmetry: $x = 4$ | # | Equation | P | q | Vertex (p,q) | Line of Symmetry (x=p) | Number of x-intercepts | | :---: | :---------------: | :---: | :---: | :---------------- : | :-------------------------: | :------------------------: | | 1 | $ y = -(x + 1)^2 + 3$ | -1 | 3 | (-1, 3), max | x = -1 | 2 | | 2 | $ y = -(x-2)^2 - 5$ | 2 | -5 | (2, -5), max | x = 2 | 0 | | 3 | $ y = (x + 4)^2$ | -4 | 0 | (-4, 0), min | x = -4 | 1 | | 4 | $ y = (x - 5)^2 - 1$ | 5 | -1 | (5, -1), min | x = 5 | 2 | | 5 | $ y = -x^2 + 3$ | 0 | 3 | (0, 3), max | x = 0 | 2 | | 6 | $ y = -(x - \frac{3}{4})^2 + \frac{7}{16}$ | $\frac{3}{4}$ | $\frac{7}{16}$ | ($\frac{3}{4}$, $\frac{7}{16}$), max | $x = \frac{3}{4}$ | 2 | ## Sketch the graph of $y = (x-1)^2 - 4$ a) Opening: **up** b) Vertex: **V(1, -4)** c) y-intercept: let $x = 0$ * $y = (0 - 1)^2 - 4$ * $y = -3$ * Therefore, the **y-intercept** is (-3). d) Let $y = 0$ * $0 = (x - 1)^2 -4$ * $±√4 = (x - 1)^2$ * $±2 = x - 1$ * $1 ± 2 = x$ * $x = 3, x = -1$ * Therefore, the **x-intercepts** are 3 and -1. e) Line of symmetry: $x = 1$ ## Sketch $y = -(x + 5)^2 + 1$ a) Opening: **downward** b) Vertex: **(-5, 1)** c) y-intercept: let $x = 0$ * $y = -(0 + 5)^2 + 1$ * $y = - 25 + 1$ * $y = -24$ * Therefore, the **y-intercept** is -24. d) x-intercept: let $y = 0$ * $0 = -(x + 5)^2 + 1$ * $(x + 5)^2 = 1$ * $x + 5 = ±1$ * $x = -5 ±1$ * $x = -4, x = -6$ * Therefore, the **x-intercepts** are -4 and -6.