Find Quadratic Functions Given Their Graphs

Detailed Guide with Vertex, Intercept, and Symmetry Axis Methods

To find the equation of a quadratic function from its graph, we identify key features such as the vertex, intercepts, or the axis of symmetry. Depending on the information available, we use either the vertex form or the factored form to determine the final function.

Key Quadratic Forms

General Form: \[ f(x) = ax^2 + bx + c \]

Vertex Form (Standard Form): \[ f(x) = a(x - h)^2 + k \]

Where $(h, k)$ are the coordinates of the vertex. The relationship between forms is given by:

\[ h = \frac{-b}{2a}, \quad k = f(h) = c - \frac{b^2}{4a} \]

Method 1: Knowing Vertex and a Point

Example 1

Find the quadratic function $f$ whose graph is shown below:

Graph of quadratic function with vertex at origin
View Solution

1. Identify the vertex from the graph: $(h, k) = (0, 2)$.

2. Set up the vertex form: \[ f(x) = a(x - 0)^2 + 2 = ax^2 + 2 \]

3. Use point $(1, 3)$ to solve for $a$: \[ f(1) = a(1)^2 + 2 = 3 \Rightarrow a = 1 \]

Final Function: \[ f(x) = x^2 + 2 \]

Example 2: Downward Parabola

Find the function $g$ and evaluate $g(-3)$:

Downward parabola graph
View Solution

1. Vertex: $(0, -1)$. Vertex form: $g(x) = ax^2 - 1$.

2. Use point $(1, -2)$: \[ a(1)^2 - 1 = -2 \Rightarrow a = -1 \]

3. Equation: $g(x) = -x^2 - 1$.

4. Evaluate $g(-3)$: \[ g(-3) = -(-3)^2 - 1 = -9 - 1 = -10 \]

Example 3: Offset Vertex

Find the function $l$ and calculate its x-intercepts:

Parabola with vertex (2,1)
View Solution

1. Vertex $(2, 1) \Rightarrow l(x) = a(x - 2)^2 + 1$.

2. Use y-intercept $(0, -7)$: \[ a(0 - 2)^2 + 1 = -7 \Rightarrow 4a = -8 \Rightarrow a = -2 \]

3. Find x-intercepts (set $l(x) = 0$): \[ -2(x - 2)^2 + 1 = 0 \Rightarrow (x - 2)^2 = 1/2 \]

\[ x = 2 \pm \sqrt{1/2} \]

Method 2: Knowing x and y Intercepts

Example 4

Find the function $s$ in standard form:

Parabola with two x-intercepts
View Solution

1. Use factored form with intercepts $(-1, 0)$ and $(2, 0)$: \[ s(x) = a(x + 1)(x - 2) \]

2. Use y-intercept $(0, -4)$: \[ a(0 + 1)(0 - 2) = -4 \Rightarrow -2a = -4 \Rightarrow a = 2 \]

3. Standard form: \[ s(x) = 2(x^2 - x - 2) = 2x^2 - 2x - 4 \]

Method 3: Knowing Axis and Two Points

Example 5

Find $m(x)$ knowing the axis of symmetry is $x = -3$:

Parabola with axis of symmetry indicated
View Solution

1. Vertex form: $m(x) = a(x + 3)^2 + k$.

2. Create a system of equations using points $(-5, 0)$ and $(-2, -1.5)$:

\[ \begin{cases} a(-5 + 3)^2 + k = 0 \Rightarrow 4a + k = 0 \\ a(-2 + 3)^2 + k = -1.5 \Rightarrow a + k = -1.5 \end{cases} \]

3. Subtracting the equations: $3a = 1.5 \Rightarrow a = 0.5$. Then $k = -2$.

4. Standard form: \[ m(x) = 0.5(x + 3)^2 - 2 = 0.5x^2 + 3x + 2.5 \]

Method 4: Knowing Three Points

Example 6

Determine $w(x)$ in standard form:

General parabola with three points marked
View Solution

Using $w(x) = ax^2 + bx + c$:

  • Point $(0, -1/6) \Rightarrow c = -1/6$.
  • Point $(1, 0) \Rightarrow a + b - 1/6 = 0 \Rightarrow a + b = 1/6$.
  • Point $(3, 10/3) \Rightarrow 9a + 3b - 1/6 = 10/3 \Rightarrow 9a + 3b = 7/2$.

Solving the system: $a = 1/2, b = -1/3$.

Final Function: \[ w(x) = \frac{1}{2}x^2 - \frac{1}{3}x - \frac{1}{6} \]

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