Answer:
x-intercept = -6 y-intercept = 9
Step-by-step explanation:
Step-by-step explanation:
h(x) = (x-11)(x+3)
= x^2 + 3 - 11x - 33
= x^2 - 11x -30
Answer:
The other 3 vertices would be (10,5) , (5,0) , and (0,5)
Step-by-step explanation:
Answer:
y=1/6x-6
Explanation:
What we know:
- Equation for line S: y=-6x+2
- Line T is perpendicular to line S
- Line T passes through the point (-6,-7)
To find the answer, we would need to understand two things, the first being what exactly is going on with the equation: y=-6x+2.
It’s organized in slope-intercept form, which is y=mx+b. Here’s what the variables mean:
Y=the y variable
M=the slope
X=the x variable
B=the y-intercept (the point on the line that crosses the y axis)
Knowing this, we can look at the equation for line S again: y=-6x+2. -6 is the slope of line S and 2 is the y-intercept of line S.
The second thing we need to understand is that the slope of a second line that is perpendicular to the first line is equal to the NEGATIVE RECIPROCAL of the slope of the first line.
Reciprocal: the “flip” of a fraction where the denominator and the numerator switch places
So, for example, let’s say the slope of line A=6. The slope of the line perpendicular to line A would be -1/6.
Let’s try another example: the slope of line B=-4/5. The slope of the line perpendicular to line B would be 5/4 (because two negatives makes a positive)
And here, for your question, the slope of line S=-6. Therefore, the slope of line T would be 1/6.
Awesome! We found the slope, which means so far, the equation of line T looks like this:
y=1/6x+b
All we need now is the y-intercept, which is “b” in the equation y=mx+b.
To do this, we plug in the point that the question provided us, (-6,-7), and solve. We would plug in -7 as the y variable and -6 as the x variable.
y=1/6x+b
-7=1/6(-6)+b
1) multiply 1/6 and -6
-7=-1+b
2) add 1 to both sides of the equation to isolate b
-6=b
Aha! There we have it, -6 is the y-intercept of line T.
Now, we have completed our equation:
y=1/6x-6
I hope this helps! Please comment if you have any questions.
