Answer:
A. x = 1 + √11 or x = 1 − √11
B. x = 1 + √3 or x = 1 − √3
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Answer:......................................
Peter.
Answer:
![y=4x^2-8x-4](https://tex.z-dn.net/?f=y%3D4x%5E2-8x-4)
Step-by-step explanation:
<u>Quadratic Function</u>
The quadratic function can be expressed in the following form:
![y=a(x-x_1)(x-x_2)](https://tex.z-dn.net/?f=y%3Da%28x-x_1%29%28x-x_2%29)
Where a is a real number different from 0, and x1, x2 are the roots or zeroes of the function.
From the conditions stated in the problem, we know
x_1=1+\sqrt{2}, \ x_1=1-\sqrt{2}
Substitute in the general formula above:
![y=a[x-(1+\sqrt{2})][x-(1-\sqrt{2})]](https://tex.z-dn.net/?f=y%3Da%5Bx-%281%2B%5Csqrt%7B2%7D%29%5D%5Bx-%281-%5Csqrt%7B2%7D%29%5D)
Operate the indicated product
![y=a(x^2-2x-1)](https://tex.z-dn.net/?f=y%3Da%28x%5E2-2x-1%29)
To find the value of a, we use the y-intercept which is the value of y when x=0, thus
![y=a(0^2-2(0)-1)=-4](https://tex.z-dn.net/?f=y%3Da%280%5E2-2%280%29-1%29%3D-4)
It follows that
![a=4](https://tex.z-dn.net/?f=a%3D4)
Thus, the required quadratic function is
![y=4(x^2-2x-1)](https://tex.z-dn.net/?f=y%3D4%28x%5E2-2x-1%29)
Or, equivalently
![\boxed{y=4x^2-8x-4}](https://tex.z-dn.net/?f=%5Cboxed%7By%3D4x%5E2-8x-4%7D)
Answer:
y-4 = (-1/3)(x+1)
Step-by-step explanation:
We need to identify the equation in point slope form.
The standard equation of point slope form is:
(y-y₁) = m (x-x₁)
where m is the slope and x₁ and y₁ are the points
We are given point(4,-1) so,
x₁=4 and y₁=-1
And a perpendicular line: y =3x+5
Which is equal to y = mx+b
where m is slope so, slope m = 3
Since the line is perpendicular, so the slope in negative inverse of actual slope that m = -1/m
i.e, m = -1/3
So, the equation in point slope form is:
y-(-1) = (-1/3)(x-4)
=> y+1 = (-1/3)(x-4)
Answer:
A)-4,-5
B)4,5
Step-by-step explanation: