Simplify the polynomial. Then find its value for the given value of the variable.
1 answer:
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We have that
<span>A (-8, -2) and B(16,6)
step 1
find the distance AB in the x coordinates
dABx=(16-(-8))-----> 24 units
step 2
find coordinate x of P (Px)
Px=Ax+(3/5)*dABx------> Px=(-8)+(3/5)*24----> 6.4
step 3
F</span>ind the distance AB in the y coordinates
dABy=(6-(-2))-----> 8 units
step 4
find coordinate y of P (Py)
Py=Ay+(3/5)*dABy------> Py=(-2)+(3/5)*8----> 2.8
the coordinates of P are (6.4,2.8)
see the attached figure
<span>A (-8, -2) and B(16,6)
step 1
find the distance AB in the x coordinates
dABx=(16-(-8))-----> 24 units
step 2
find coordinate x of P (Px)
Px=Ax+(3/5)*dABx------> Px=(-8)+(3/5)*24----> 6.4
step 3
F</span>ind the distance AB in the y coordinates
dABy=(6-(-2))-----> 8 units
step 4
find coordinate y of P (Py)
Py=Ay+(3/5)*dABy------> Py=(-2)+(3/5)*8----> 2.8
the coordinates of P are (6.4,2.8)
see the attached figure
Answer:
Step-by-step explanation:
We have the function:
And we want to find x=d for which the minimum/maximum value will occur.
Notice that our function is a quadratic in factored form.
Remember that the minimum/maximum value always occurs at the vertex point.
And remember that the x-coordinate of the vertex is the axis of symmetry.
Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.
Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.
Our factors are <em>(x-2) </em>and <em>(x-1)</em>.
Therefore, our roots/zeros are <em>x=1, 2</em>.
So, the average of them are:
Therefore, regardless of the value of <em>a</em>, the minimum/maximum value will occur at <em>x=d=1.5</em>.
Alternative Method:
Of course, we can also expand to confirm our answer. So:
The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.
And the formula for the vertex is:
So, we will substitute <em>-3a</em> for <em>b</em> and <em>a</em> for <em>a</em>. This yields:
Confirming our answer.