The function that has a vertex on the y-axis is f(x) = (x - 2)(x + 2)
<h3>How to determine the function?</h3>
For a function to have its vertex on the y-axis, then the coordinate of the vertex must be:
(h,k) = (0,y)
A quadratic function is represented as:
f(x) = (x - h)^2 + k
So, we have:
f(x) = (x - 0)^2 + k
Evaluate
f(x) = x^2 + k
From the list of options, we have:
f(x) = (x - 2)(x + 2)
Expand
f(x) = x^2 - 4
Hence, the function that has a vertex on the y-axis is f(x) = (x - 2)(x + 2)
Read more about vertex at:
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The roots of the polynomial <span><span>x^3 </span>− 2<span>x^2 </span>− 4x + 2</span> are:
<span><span>x1 </span>= 0.42801</span>
<span><span>x2 </span>= −1.51414</span>
<span><span>x3 </span>= 3.08613</span>
x1 and x2 are in the desired interval [-2, 2]
f'(x) = 3x^2 - 4x - 4
so we have:
3x^2 - 4x - 4 = 0
<span>x = ( 4 +- </span><span>√(16 + 48) </span>)/6
x_1 = -4/6 = -0.66
x_ 2 = 2
According to Rolle's theorem, we have one point in between:
x1 = 0.42801 and x2 = −1.51414
where f'(x) = 0, and that is <span>x_1 = -0.66</span>
so we see that Rolle's theorem holds in our function.
Answer:
x = 6
y = - 14
Step-by-step explanation:
- -2x × 2 + 10 = - 2x - 2
- x = 6
- y = -2 × 6 - 2 = - 14
Answer:
N
Step-by-step explanation:
