HELP please!!! For two figures to be congruent their corresponding sides and exterior angles must be equal
1 answer:
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Answer:
<em>The coefficient of the squared expression in the parabolas equation will be 5.</em>
Step-by-step explanation:
<u>The vertex form of parabola</u> is: , where
is the vertex point and
is the coefficient of
term.
The vertex is given as . That means,
and
So, the vertex form will be:
Given that, when the x-value is 4, the y-value is 3. So, plugging these values into the above equation, we will get.....
Thus, the coefficient of the squared expression in the parabolas equation will be 5.
Answer:
(A) -3 ≤ x ≤ 1
Step-by-step explanation:
The given function is presented as follows;
h(x) = x² - 1
From the given function, the coefficient of the quadratic term is positive, and therefore, the function is U shaped and has a minimum value, with the slope on the interval to the left of <em>h</em> having a negative rate of change;
The minimum value of h(x) is found as follows;
At the minimum of h(x), h'(x) = d(h(x)/dx = d(x² - 1)/dx = 2·x = 0
∴ x = 0/2 = 0 at the minimum
Therefore, the function is symmetrical about the point where x = 0
The average rate of change over an interval is given by the change in 'y' and x-values over the end-point in the interval, which is the slope of a straight line drawn between the points
The average rate of change will be negative where the y-value of the left boundary of the interval is higher than the y-value of the right boundary of the interval, such that the line formed by joining the endpoints of the interval slope downwards from left to right
The distance from the x-value of left boundary of the interval that would have a negative slope from x = 0 will be more than the distance of the x-value of the right boundary of the interval
Therefore, the interval over which <em>h</em> has a negative rate of change is -3 ≤ x ≤ 1
