Find the slope of the line that passes though (10, 3) and (6, 2)
2 answers:
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Answer:
Step-by-step explanation:
<u>Part A: the x-intercepts are -3/2 and 5/2 </u>
- 4
- 7x - 15 -----> (2x + 3) (2x - 5)
2x + 3 = 0 --------> 2x = -3 ----------> x =
2x - 5 = 0 ---------> 2x = 5 -----------> x =
<u>Part B: the parabola is a minimum, the vertex is (7/8,-289/16) </u>
f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
a<0 then the parabola opens down and the vertex is a max
- a = 4, it is a minimum
the x value of the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a))
f(x) = 4 - 7x - 15
a = 4
b = -7
-b/2a=-(7)/(2*4)=7/8
f(7/8) = 4 - 7(7/8) - 15
f(7/8) = 49/16 - 49/8 - 15
f(7/8) = -289/16
- the vertex is (7/8,-289/16)
<u>Part C: </u>
- the vertex is minimum and the graph goes through the x intercepts
- plug in x values to get the y value (EX: choose 0 for x, and you'll get -15 for y, so you would plot the point at (0, -15). you could plug in 1 for x, and get -18 for y, plotting the point (1, -18) )