The function f is given in three equivalent forms. Which form most quickly reveals the zeros (or 'roots") of the function?

Mathematics

1 answer:

shepuryov [24]3 years ago

3 0

9514 1404 393

Answer:

A, 2

Step-by-step explanation:

In the factored form, the zeros are the values of x that make a factor zero. That is, one of the zeros is ...

x -2 = 0 ⇒ x = 2

To find the zeros from the other forms, more work must be done. In the standard form, one of the usual solution methods must be used, possibly including putting the equation into factored form, completing the square, or using the quadratic formula. In vertex form, a root must be taken, then those values adjusted by translation to the axis of symmetry.

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Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :