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Chapter 6: Problem 39

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$y=\frac{1}{4}\left(x^{2}-2 x+5\right)$$

Short Answer

Expert verified
The vertex of the parabola is located at (2, 0.25), the focus is at (2, 1.25), and the directrix is the line \(y = -0.75\).

Step by step solution

01

Finding the vertex

The vertex of a parabola \((h, k)\) can be found by using the formula \(h = -\frac{b}{2a}\), where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation \(ax^2+bx+c\). For the given equation \(\frac{1}{4}(x^2-2x+5)\), \(a = \frac{1}{4}\), \(b = -2\), and \(c = 5\). Plug these values into the h formula: \(h = -(-2)/(2*(\frac{1}{4})) = 2\) and substituting x = h in given equation, we get \(k = \frac{1}{4}*(2^2-2*2+5) = \frac{1}{4}*1 = 0.25\). So, the vertex is (2, 0.25).
02

Finding the focus

The focus of the parabola is found by using the formula \(f = \frac{1}{4a}\). Here a is coefficient of \(x^2\), which is \(\frac{1}{4}\), So the focus will be \(f = \frac{1}{4(\frac{1}{4})} = 1\). Since the parabola is opening upwards, the coordinate of the focus would be \((h, k+f) => (2, 0.25+1) => (2, 1.25)\).
03

Finding the directrix

The directrix of the parabola is a line opposite to the focus and equidistant from the vertex. So the directrix is the line \(y=k-f => y=0.25-1=y=-0.75\).
04

Sketching the parabola

To sketch the parabola: mark the point (2, 0.25) as the vertex, plot the focus at (2, 1.25), and draw the line \(y = -0.75\) as the directrix. The parabola will open upwards, because \(a\), the coefficient of \(x^2\), is positive.

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