Answer:
Step-by-step explanation:
We want to find an equation of a line that's perpendicular to x=1 that also passes through the point (8,-9).
Note that x=1 is a <em>vertical line </em>since x is 1 no matter what y is.
This means that if our new line is perpendicular to the old, then it must be a <em>horizontal line</em>.
So, since we have a horizontal line, then our equation must be our y-value of our point.
Our y-coordinate of our point (8,-9) is -9.
Therefore, our equation is:
And this is in standard form.
And we're done!
<em>Refer to the attachment for solution and below for steps!!~</em>
Answer:
P(x) =x² + 3x + 2
Step-by-step explanation:
You have two zeros, so the polynomial is a quadratic.
The general form of a quadratic is
P(x) = ax² + bx + c
Sum of zeros = -b
Product of zeros = c
The polynomial becomes
P(x) = x² - (sum of zeros)x + (product of zeros)
Sum of zeros = -3 and product of zeros = 2, so
P(x) = x² + 3x + 2
The graph of P(x) (below) shows that the zeros are at -2 and -1
Sum of zeros = -2 + (-1) = -3
Product of zeros = -2 × (-1) = 2
