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Gnoma [55]
3 years ago
8

When graphing y = 2x2 + 35x + 75, which viewing window would allow you to see all of the intercepts and the minimum as closely a

s possible?
Mathematics
1 answer:
djverab [1.8K]3 years ago
5 0

The intercept with the y-axes occurs when x = 0

f(0) = 2*02 + 35*0 + 75 = 0 + 0 + 75 = 75

So intercept with x axes is the point (0,75)

The intercept with the x-axes occurs when y=0

2x2 + 35x + 75 = 0

<span>As there’s no direct way to find x, we can use the <span>Bhaskara formula:</span></span>

x=(-b+- square root (b2-4ac))/(2a)

This formula is for: ax2 +bx + c = 0

So:

a=2

b=35

c=75

x=(-35+- square root (35^2-4*2*75))/(2*2)

So x can be -2.5 or -15

So intercept with y axes are the points (-2.5,0) and (-15,0)

And there is also a formula to find the mínimum. The mínimum is always the vertex:

Xv = -b/(2*a) = -35/(2*2) = -8.75

Yv = f(Xv) = f(-8.75) = 2(-8.75)^2 + 35(-8.75) <span> + 75 = </span>-78.125

So the minimum is the point (-8.75, -78.125)

You can easily verify all these data with a graphic software such as GeoGebra.

Here is the complete list of all the points we need to see:

(0,75); (-2.5,0) ; (-15,0), (-8.75, -78.125)

So the mínimum X has to be -15 and the máximum X has to be 0

and the mínimum Y has to be -78.125 and the máximum Y has to be 75

So this leads to a different scale for X and for Y.


Window x = [ -15, 0] y = [ -78.125, 75]
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let <A, <B,<C are three angles in a triangle.

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This implies that,

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If f(x)=x^3-x+2, then (f^-1)'(2)
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f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

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