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5. Find the quadratic function whose graph has vertex at (1,2) and passes through (3,4).

5. Find the quadratic function whose graph has vertex at (1,2) and passes through (3,4).

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem 5** Find the quadratic function whose graph has a vertex at \((1,2)\) and passes through \((3,4)\). ### Solution To find the quadratic function, we can use the vertex form of a quadratic equation: \[ y = a(x - h)^2 + k \] where \((h, k)\) is the vertex of the parabola. Given the vertex \((1, 2)\), we can substitute \(h = 1\) and \(k = 2\) into the vertex form equation: \[ y = a(x - 1)^2 + 2 \] Next, we use the point \((3,4)\) which lies on the parabola to determine the value of \(a\). Substitute \(x = 3\) and \(y = 4\) into the equation: \[ 4 = a(3 - 1)^2 + 2 \] \[ 4 = a(2)^2 + 2 \] \[ 4 = 4a + 2 \] \[ 4 - 2 = 4a \] \[ 2 = 4a \] \[ a = \frac{1}{2} \] Thus, the quadratic function is: \[ y = \frac{1}{2}(x - 1)^2 + 2 \] ### Graphical Explanation The equation represents a parabola that opens upwards (since \(a > 0\)). The vertex of the parabola is at \((1, 2)\), and it passes through the point \((3, 4)\). The vertex form directly tells us where the vertex is located and how wide or narrow the parabola is. The coefficient \(a\) determines the "width" or "steepness" of the parabola. This quadratic function can also be expanded into standard form if preferred: \[ y = \frac{1}{2}(x^2 - 2x + 1) + 2 \] \[ y = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{2} + 2 \] \[ y = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{5}{2} \] These steps highlight how to derive the quadratic function from a vertex and an additional
Transcribed Image Text:**Problem 5** Find the quadratic function whose graph has a vertex at \((1,2)\) and passes through \((3,4)\). ### Solution To find the quadratic function, we can use the vertex form of a quadratic equation: \[ y = a(x - h)^2 + k \] where \((h, k)\) is the vertex of the parabola. Given the vertex \((1, 2)\), we can substitute \(h = 1\) and \(k = 2\) into the vertex form equation: \[ y = a(x - 1)^2 + 2 \] Next, we use the point \((3,4)\) which lies on the parabola to determine the value of \(a\). Substitute \(x = 3\) and \(y = 4\) into the equation: \[ 4 = a(3 - 1)^2 + 2 \] \[ 4 = a(2)^2 + 2 \] \[ 4 = 4a + 2 \] \[ 4 - 2 = 4a \] \[ 2 = 4a \] \[ a = \frac{1}{2} \] Thus, the quadratic function is: \[ y = \frac{1}{2}(x - 1)^2 + 2 \] ### Graphical Explanation The equation represents a parabola that opens upwards (since \(a > 0\)). The vertex of the parabola is at \((1, 2)\), and it passes through the point \((3, 4)\). The vertex form directly tells us where the vertex is located and how wide or narrow the parabola is. The coefficient \(a\) determines the "width" or "steepness" of the parabola. This quadratic function can also be expanded into standard form if preferred: \[ y = \frac{1}{2}(x^2 - 2x + 1) + 2 \] \[ y = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{2} + 2 \] \[ y = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{5}{2} \] These steps highlight how to derive the quadratic function from a vertex and an additional
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