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3 years ago

How would you find the x-intercept of f(x)=(x-4)^2-3?

1 answer:

8 0

The x -intercepts occur when f(x) =0
So,
(x-4)^2 = x^2 -8x + 16 -3 =0
x^2 -8x + 13 =0
Solving by the quadratic formula
x1 = 5.7321
x2 = 2.2679

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Since, $\tan x=\frac{o p p}{a d j}$

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To find the hypotenuse, let us use the pythagoras theorem,

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Now, we can find the value of sin x and cos x.

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$\cos x=\frac{a d j}{h y p}=\frac{5}{13}$

Now, substituting these values in the formula for sin 2x, we get,

$\begin{aligned}\sin (2 x) &=2 \sin x \cos x \\&=2\left(\frac{12}{13}\right)\left(\frac{5}{13}\right) \\&=\frac{120}{169}\end{aligned}$

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