Find the midpoint of the line segment with the given endpoint (-3,4) (-2,-8)
1 answer:
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To determine the maximum value of a quadratic function opening downwards, we are going to find the vertex; then the y-value of the vertex will be our maximum.
To find the vertex (h,k) (where h=x-coordinate and k=y-coordinate) of a quadratic function of the form
we'll use the vertex formula: 
, and then we are going to replace that value in our original function to find k.
So, in our function
, 
and 
.
Lets replace those values in our vertex formula:
Now that we know the x-coordinate of our vertex, lets replace it in the original function, to get the y-coordinate:
We just prove that the vertex of
is (2,1), and for the graph we can tell that the vertex of 
is (-2,4). The only thing left is compare their y-coordinates to determine w<span>hich one has the greater maximum value. Since 4>1, we can conclude that </span> has the greater maximum.
To find the vertex (h,k) (where h=x-coordinate and k=y-coordinate) of a quadratic function of the form
So, in our function
Lets replace those values in our vertex formula:
Now that we know the x-coordinate of our vertex, lets replace it in the original function, to get the y-coordinate:
We just prove that the vertex of
ANSWER
EXPLANATION
Since the altitude, |GJ| bisects the base of the triangle, it follows that, the triangle is an isosceles triangle.
This implies that,.
In terms of x, we have the equation.
We now group like terms to obtain,
This implies that,
Let us divide through by 2 to get,
We can see from the diagram that,
We substitute x=5 to get,
This simplifies to,
EXPLANATION
Since the altitude, |GJ| bisects the base of the triangle, it follows that, the triangle is an isosceles triangle.
This implies that,.
In terms of x, we have the equation.
We now group like terms to obtain,
This implies that,
Let us divide through by 2 to get,
We can see from the diagram that,
We substitute x=5 to get,
This simplifies to,