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Marina CMI [18]
2 years ago
11

8. The graph of f(x) = 2x² - 5x is shown. Amanda

Mathematics
1 answer:
snow_lady [41]2 years ago
5 0

Answer: Amanda is not correct

The vertex is at (1.25, -3.125)

=========================================================

Reason:

The x intercepts are 0 and 2.5

This is where the graph crosses the x axis. If we plugged in either x = 0 or x = 2.5, then we'd get f(x) = 0.

Find the midpoint of those values mentioned. Add them up, and divide by 2 to get the midpoint to be (0+2.5)/2 = 1.25

This is the x coordinate of the vertex.

Then plug that into the function to find the corresponding y coordinate of the vertex.

f(x) = 2x^2 - 5x

f(1.25) = 2(1.25)^2 - 5(1.25)

f(1.25) = -3.125

The vertex is located at (1.25, -3.125)

Amanda's x coordinate is correct, but the y coordinate isn't. Though her y coordinate isn't too far off.

I think I can see why she went with y = -3 since the lowest part of the graph is around this area. However, the lowest part of the parabola dips a tiny bit below y = -3. This is why it's better to use the equation rather than solely rely on the graph alone.

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Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
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Step-by-step explanation:

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Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

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