This one looks very tricky but ill try it out
The statements true about the the function f(x) = 2x2 – x – 6 are-
- The vertex of the function is (one-quarter, negative 6 and one-eighth).
- The function has two x-intercepts.
<h3>What is vertex of parabola?</h3>
The vertex of parabola is the point at the intersection of parabola and its line of symmetry.
Now the given function is,
f(x) = 2x^2 – x – 6
Also, it is given that the vertex is located at (0.25, -6) and the parabola opens up, the function has two x-intercepts.
Comparing the given function with standard form,
f(x) = a x^2 bx + c
By comprison we get,
a = 2
b = -1
c = -6
Now, x-coordinate of vertex is given as,
x = -b/2a
put the values we get,
x = -(-1)/2*2
or, x = 1/4
Put the value of x in given function, so y-coordinate of the vertex is given as,
f(1/4) = 2(1/4)² - 1/4 - 6
= -49/6
= -6 1/8
Hence, The statements true about the the function f(x) = 2x2 – x – 6 are-
- The vertex of the function is (one-quarter, negative 6 and one-eighth).
- The function has two x-intercepts.
More about vertex :
brainly.com/question/86393
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Answer:
D.
Step-by-step explanation:
After the third step, take the square root of both sides.
Third Step:
Thus, the fourth step is:
The left side cancels:
D.
Answer:
It is the 3rd graph. The other dude lied and tried to trick me but its the 3rd graph.
Step-by-step explanation:
The 3rd graph C
Differentiating an integral removes the integral.
f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.
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