Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

what are the zeros of the function f(x)=x^2-2x-15?

Sagot :

ddk
To find the zeros of this function, we must first set the entire function equal to 0

f(x) = x² - 2x - 15 = 0

Since this is a quadratic function, we must use the quadratic formula, which is:

[tex] \frac{-b +/- \sqrt{b^{2} - 4(a)(c) } }{2a} [/tex]

Let's assign a, b, and c using our first function
x² means a = 1 (because it could be written as 1x²)
-2x means b = -2
-15 means c = -15

Now let's plug those in:

[tex] \frac{-(-2) +/- \sqrt{(-2)^{2} - 4(1)(-15) } }{2(1)} [/tex]

which simplifies to:

[tex] \frac{2 +/- \sqrt{(4 + 60} }{2} [/tex]

Simplified further:

[tex] \frac{2 +/- \sqrt{(64} }{2} [/tex]
[tex]\frac{2 +/- 8 }{2} [/tex]
And divide it by the 2 on the bottom gives us:

[tex]2 +/- 4[/tex]

2+4 = 6
2-4 = -2

So the zeros of this function are -2 and 6
boffeemadrid

The zeros of the given function is required.

The zeros of the polynomial are [tex]5,-3[/tex].

The function is

[tex]f(x)=x^2-2x-15[/tex]

[tex]a=1[/tex]

[tex]b=-2[/tex]

[tex]c=-15[/tex]

Using the quadratic formula

[tex]x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\Rightarrow x=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\times1\times \left(-15\right)}}{2\times 1}\\\Rightarrow x=\dfrac{2\pm8}{2}\\\Rightarrow x=5,-3[/tex]

Learn more:

https://brainly.com/question/11845608

https://brainly.com/question/12129448

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.