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Chapter 2: Problem 28

Write the quadratic function in standard form and sketch its graph. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). $$f(x)=x^{2}+3 x+\frac{1}{4}$$

Short Answer

Expert verified
The standard form is \(f(x)=(x+\frac{3}{2})^2 - \frac{5}{4}\). The vertex is \((-\frac{3}{2},-\frac{5}{4})\), the axis of symmetry is \(x = -\frac{3}{2}\), and the x-intercepts are \(x=-\frac{1}{2},-2\). The graph is a parabola that opens upwards with a vertex at the point \((-\frac{3}{2},-\frac{5}{4})\), and that intersects the x-axis at the points \((-\frac{1}{2},0)\) and \((-2,0)\).

Step by step solution

01

Rewrite Function in Standard Form

To write a quadratic function in standard form, complete the square. The given quadratic function can be rewritten from \(f(x)=x^2+3x+\frac{1}{4}\) to a standard form \(f(x)=a(x-h)^2+k\) by arranging the given function into \(f(x)=(x+\frac{3}{2})^2- \frac{5}{4}\)
02

Identify the Vertex

The vertex of the parabola is at \((h,k)\). From the standard form, \(h = -\frac{3}{2}\) and \(k = -\frac{5}{4}\), so the vertex is \((-\frac{3}{2},-\frac{5}{4})\)
03

Identify the Axis of Symmetry

The axis of symmetry is the line \(x=h\). In this case, the axis of symmetry is the line \(x=-\frac{3}{2}\)
04

Identify the x-intercepts

The x-intercepts are where the function equals zero. Set \(f(x) = 0\) and solve for \(x\) to find the x-intercepts. We have \(0=(x+\frac{3}{2})^2 - \frac{5}{4}\), which gives \(x=-\frac{1}{2},-2\)
05

Sketch the Graph

Plot the vertex, axis of symmetry, and x-intercepts on a graph. Sketch the parabola using these points and the fact that parabolas are symmetric to their axis of symmetry.

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