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2.2: Equations of Lines and Quadratic Functions

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    231
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    1. Parallel Lines and Perpendicular Lines

    We define parallel and perpendicular lines as follows:

    \[m_1 = -\dfrac{1}{m_2}.\]

    \[\begin{align} y - 4 &= \dfrac{3}{2}(x - 1) \\ y &= \dfrac{3}{2} x +\dfrac{5}{2}. \end{align}\]

    \[x = constant.\]

    Find the equation of the line through the point \((5,4)\) that is vertical

    Solution

    We know that the form of the equation is:

    \[x = constant\]

    Since it passes through \((5,4)\) which as x-coordinate 5 that constant must be 5. We conclude that the equation of the line is

    \[x = 5.\]

    To interactively investigate equations of lines go to

    mathcsjava.emporia.edu/~greenlar/ParPerp/ParPerp.html

    2. Quadratic Functions in Standard Form

    Recall what the function

    \[y = x^2\]

    looks like. It is a parabola with vertex at the origin. We can use shifting techniques to graph

    \[y= 3(x - 2)^2+ 4.\]

    We see that the vertex is shifted to the right by 2 units and up 4 units. There is vertical stretching by a factor of three. In general, we say that a quadratic is in standard form if it looks like:

    \[y=a(x-h)^2+k.\]

    Here the \(h\) represents the horizontal shift, the \(k\) represents the vertical shift, and the \(a\) represents the stretching factor. If \(a\) is negative the parabola is concave down (looks like a frown).

    To put a quadratic in standard form, we complete the square.

    \[y = 2(x - 2)^2- 6.\]

    3. Applications:

    Newton's law of motion:

    Newton discovered that if an object is thrown from an initial height of \(s_0\) feet with an initial velocity of \(v_0\) feet per second then the position of the function of the object is

    \[s = -16t^2+ v_0t + s_0\]

    \[t = 1.74 \; \text{ and } \; t = -1.74\]

    Notice that a negative time means is not the solution, hence the ball hits the ground after 1.74 seconds.

    To find out how high the ball will go, we are after the \(s\) coordinate of the vertex. We put the equation into standard form

    Now we can say that the ball will reach a height of 9.77 feet, \(\frac{25}{32}\) seconds after it was released.

    To find out when the ball will be higher than 7 feet we solve

    \[7 < -16t^2 + 25t + 5\]

    or

    \[0 < -16t^2 + 25t - 2\]

    Using the quadratic formula we find the roots at

    \[t = 0.085 \;\;\; \text{and} \;\;\; t = 1.48\]

    So that the ball will be higher than 7 feet between 0.085 seconds and 1.48 seconds.

    1. Factor the leading coefficient:

    \[y = -16(x^2 - \dfrac{25}{16} x - \dfrac{5}{16})\]

    2. Calculate

    \[\begin{align} -\dfrac{b}{2}&=-(\dfrac{25}{16}) \\ &=-\dfrac{25}{32} \end{align}\]

    3. Square the solution above:

    \[(-\dfrac{25}{32})^2 = 0.61\]

    4. Add and subtract answer from part three inside parentheses:

    \[y = -16(x^2 - \dfrac{25}{16} x + 0.61 - 0.61 + 5/16) \]

    5. Regroup:

    \[y = -16[(x^2 - \dfrac{25}{16}x + 0.61 ) - 0.61]\]

    6. Factor the inner parentheses using part two as a hint:

    \[y = -16[(x - \dfrac{25}{32})^2 - 0.61]\]

    7. Multiply out the outer constant:

    \[y = -16(x - \dfrac{25}{32})^2 + 9.77\]

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.


    This page titled 2.2: Equations of Lines and Quadratic Functions is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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