Fall 2024 The graph of a quadratic function is given.Determine the function's equation. 1)

To determine the equation of a quadratic function from its graph, we need to identify key features such as the vertex, axis of symmetry, and any points through which the graph passes. The general form of a quadratic function is:<br /><br />\[ f(x) = ax^2 + bx + c \]<br /><br />Alternatively, if we know the vertex \((h, k)\), we can use the vertex form:<br /><br />\[ f(x) = a(x - h)^2 + k \]<br /><br />Let's go through the steps to find the equation:<br /><br />1. **Identify the Vertex**: The vertex form of a quadratic function is useful if we know the vertex. Suppose the vertex is at \((h, k)\).<br /><br />2. **Determine the Value of \(a\)**: To find \(a\), we need another point on the graph. Let's say the graph passes through the point \((x_1, y_1)\).<br /><br />3. **Substitute the Values**: Substitute the vertex and the point into the vertex form to solve for \(a\).<br /><br />4. **Write the Equation**: Once \(a\) is determined, write the full equation in vertex form or convert it to standard form if needed.<br /><br />### Example<br /><br />Suppose the graph has a vertex at \((2, 3)\) and passes through the point \((4, 7)\).<br /><br />1. **Vertex Form**: Start with the vertex form of the quadratic equation:<br /> \[ f(x) = a(x - 2)^2 + 3 \]<br /><br />2. **Substitute the Point (4, 7)**:<br /> \[ 7 = a(4 - 2)^2 + 3 \]<br /> \[ 7 = a(2)^2 + 3 \]<br /> \[ 7 = 4a + 3 \]<br /><br />3. **Solve for \(a\)**:<br /> \[ 7 - 3 = 4a \]<br /> \[ 4 = 4a \]<br /> \[ a = 1 \]<br /><br />4. **Write the Equation**:<br /> \[ f(x) = 1(x - 2)^2 + 3 \]<br /> \[ f(x) = (x - 2)^2 + 3 \]<br /><br />If you prefer the standard form, expand the vertex form:<br />\[ f(x) = (x - 2)^2 + 3 \]<br />\[ f(x) = x^2 - 4x + 4 + 3 \]<br />\[ f(x) = x^2 - 4x + 7 \]<br /><br />Thus, the equation of the quadratic function is:<br />\[ f(x) = x^2 - 4x + 7 \]<br /><br />If you provide specific details about the graph, such as the vertex and another point, I can give you a more precise equation.