Answer:
49/8 is the value of k
Step-by-step explanation:
We have the system
x = -2y^2 - 3y + 5
x=k
We want to find k such that the system intersects once.
If we substitute the second into the first giving us k=-2y^2-3y+5 we should see we have a quadratic equation in terms of variable y.
This equation has one solution when it's discriminant is 0.
Let's first rewrite the equation in standard form.
Subtracting k on both sides gives
0=-2y^2-3y+5-k
The discriminant can be found by evaluating
b^2-4ac.
Upon comparing 0=-2y^2-3y+5-k to 0=ax^2+bx+c, we see that
a=-2, b=-3, and c=5-k.
So we want to solve the following equation for k:
(-3)^2-4(-2)(5-k)=0
9+8(5-k)=0
Distribute:
9+40-8k=0
49-8k=0
Add 8k on both sides:
49=8k
Divide both sides by 8"
49/8=k
<span>A(t) = −(t − 8)2 + 535
You would like to find the maximum of this function:</span> −(t − 8)^2<span>This part is always negative or zero as a number squared cannot be negative and you multiply by -1: Thus the maximum of this part MAX:</span>−(t − 8)^2=0
<span>The max will be when t=8 and its value is 535
</span>
