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Chapter 2: Problem 26

Write the quadratic function in standard form and sketch its graph. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). $$f(x)=x^{2}-30 x+225$$

Short Answer

Expert verified
The standard form of the function is \(f(x) = (x - 15)^2\). The vertex is at \((15, 0)\), the axis of symmetry is \(x = 15\) and the x-intercept is \(x = 15\).

Step by step solution

01

Write the function in standard form.

To write the function in standard form, complete the square. First, group the \(x\) terms together: \(f(x) = (x^2 - 30x) + 225\). Then, factor \(a = 1\) out of the \(x\) terms, which leaves the expression unchanged. Find \((b/2)^2\) where \(b = -30\), which equals \(225\). Add and subtract this inside the parentheses: \(f(x) = [(x - 15)^2 - 225] + 225\). Then simplify to obtain the standard form: \(f(x) = (x - 15)^2\).
02

Identify the Vertex and Axis of Symmetry.

When a quadratic function is given in standard form \(f(x) = a(x - h)^2 + k\), the vertex of the parabola is the point \((h, k)\). From \(f(x) = (x - 15)^2\), identify the vertex as \((15, 0)\). The axis of symmetry of a parabola is the vertical line \(x = h\), so here the axis of symmetry is \(x = 15\).
03

Identify the x-intercepts.

The x-intercepts of the graph are the values of \(x\) such that \(f(x) = 0\). Set \(f(x) = 0\) and solve for \(x\), that is \((x - 15)^2 = 0\). Solving this, we get \(x = 15\), so the graph touches the x-axis at \(x = 15\).
04

Sketch the Graph.

Sketch a graph with a vertex at the point \((15, 0)\), axis of symmetry at \(x = 15\) and x-intercept at \(x = 15\). This is a parabola opening upwards since the coefficient of \(x^2\) is positive.

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