Considering the definition of zeros of a function, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.
<h3>Zeros of a function</h3>
The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.
In summary, the roots or zeros of the quadratic function are those values of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.
In a quadratic function that has the form:
f(x)= ax² + bx + c
the zeros or roots are calculated by:
![x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}](https://tex.z-dn.net/?f=x1%2Cx2%3D%5Cfrac%7B-b%2B-%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D)
<h3>This case</h3>
The quadratic function is f(x) = x² + 4x +9
Being:
the zeros or roots are calculated as:
![x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}](https://tex.z-dn.net/?f=x1%3D%5Cfrac%7B-4%2B%5Csqrt%7B4%5E%7B2%7D-4x1x9%20%7D%20%7D%7B2x1%7D)
![x1=\frac{-4+\sqrt{16-36 } }{2x1}](https://tex.z-dn.net/?f=x1%3D%5Cfrac%7B-4%2B%5Csqrt%7B16-36%20%7D%20%7D%7B2x1%7D)
![x1=\frac{-4+\sqrt{-20 } }{2x1}](https://tex.z-dn.net/?f=x1%3D%5Cfrac%7B-4%2B%5Csqrt%7B-20%20%7D%20%7D%7B2x1%7D)
and
![x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}](https://tex.z-dn.net/?f=x2%3D%5Cfrac%7B-4-%5Csqrt%7B4%5E%7B2%7D-4x1x9%20%7D%20%7D%7B2x1%7D)
![x2=\frac{-4-\sqrt{16-36 } }{2x1}](https://tex.z-dn.net/?f=x2%3D%5Cfrac%7B-4-%5Csqrt%7B16-36%20%7D%20%7D%7B2x1%7D)
![x2=\frac{-4-\sqrt{-20} }{2x1}](https://tex.z-dn.net/?f=x2%3D%5Cfrac%7B-4-%5Csqrt%7B-20%7D%20%7D%7B2x1%7D)
If the content of the root is negative, the root will have no solution within the set of real numbers. Then
has no solution.
Finally, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.
Learn more about the zeros of a quadratic function:
brainly.com/question/842305
brainly.com/question/14477557
#SPJ1
Answer:
I'm pretty sure the answer to the third question is C
Step-by-step explanation:
We are solving for h(g(x)).
We already know that g(x) is equal to x^2+4.
So know, we can simplify our expression, h(g(x)), to h(x^2+4).
We also know that h(x)=1/x
we simply substitute x for x^2+4, giving us an answer of 1/(x^2+4)
Answer:
=−24 hope this helps!!
Step-by-step explanation:
I’m not completely sure but I think it’s
4 x 103 2 x 103 = 8 x 106 (btw I don’t know how to write the exponent so the ones next to the 10’s are the exponents)