A.) Determine "a" and "k" so both points are on the graph of the function.
1.) (0,1) (2,1); y=a(x-1)^2+k
2.) (1,11) (2,-19); y=a(x+1)^2+k
B.) Determine whether or not the function f(x)=0.25(2x-15)^2+15 has a mazimum or a minimum value. Then find the value.
C.) When y=2(x-3)(x+5) is written in standard form, what is the value of b?
D.) For y=3x^2-7x+5, what is the x-value of the vertex? Enter your answer as an improper fraction in simplest form.
THANK YOU SO MUCH!!!!! :)
A1
1 = a(-1)^2 + k
1 = a(1)^2 + k
----------------
-1 = 0 + 0 no solution possible
A2
11 = a(2^2) + k
-19 = a(3^2) + k
-----------------subtract
30 = -5 a
a = -6
11 = -6(4) + k
k = 35
so
y = -6(x+1)^2 + 35
B.) Determine whether or not the function f(x)=0.25(2x-15)^2+15 has a mazimum or a minimum value. Then find the value.
when x gets big + or big -, y gets big +
SO
This parabola has a minimum (holds water)
it is symmetric about the point where
2x-15 = 0
or x = 7.5
then calculate y at x = 7.5 to get the vertex
C.) When y=2(x-3)(x+5) is written in standard form, what is the value of b?
y = 2(x^2 + 2 x - 15)
y = 2 x^2 + 4 x - 30
D.) For y=3x^2-7x+5, what is the x-value of the vertex? Enter your answer as an improper fraction in simplest form.
3 x^2 - 7 x = y - 5
x^2 -(7/3) x = y/3 - 5/3
x^2-(7/3)x+(7/6)^2 = y/3 -5/3+(7/6)^2
(x-7/6)^2 = y/3 etc etc
so
when x = 7/6 we are at the vertex
A.) To determine the values of "a" and "k" so that both points are on the graph of the function, we can substitute the given points into the equation and solve for the variables.
1.) (0,1): Substitute x=0 and y=1 into the equation y=a(x-1)^2+k
1 = a(0-1)^2 + k
1 = a(1) + k
1 = a + k
2.) (2,1): Substitute x=2 and y=1 into the equation y=a(x-1)^2+k
1 = a(2-1)^2 + k
1 = a(1) + k
1 = a + k
Now we have two equations:
1 = a + k
1 = a + k
Since both equations are the same, we can combine them:
2 = 2a + 2k
Divide both sides of the equation by 2:
1 = a + k
Therefore, the values of "a" and "k" that satisfy both points are: a = 1 and k = 0.
B.) To determine whether the function has a maximum or minimum value, we need to analyze the coefficient of the squared term in the function.
The function is in the form: y = a(x-h)^2 + k
In this case, we have: y = 0.25(2x-15)^2 + 15
The coefficient of the squared term is positive (0.25), which means the parabola opens upwards and the function has a minimum value.
To find the value of this minimum, we need to determine the vertex of the parabola.
The x-coordinate of the vertex can be found using the formula: x = -b / (2a)
In this case, a = 0.25 and b = 2(-15) = -30.
x = -(-30) / (2*0.25)
x = 30 / 0.5
x = 60
To find the corresponding y-coordinate, substitute the x-value into the function:
y = 0.25(2(60)-15)^2 + 15
y = 0.25(120-15)^2 + 15
y = 0.25(105)^2 + 15
y = 0.25(11025) + 15
y = 2756.25 + 15
y = 2771.25
Therefore, the function has a minimum value of 2771.25.
C.) To write the equation y = 2(x-3)(x+5) in standard form (y = ax^2 + bx + c), we need to expand and simplify the expression.
y = 2(x-3)(x+5)
y = 2(x^2 + 5x - 3x - 15)
y = 2(x^2 + 2x - 15)
y = 2x^2 + 4x - 30
Comparing this form to the standard form, we can see that the value of b is 4.
D.) The x-value of the vertex of the function y = 3x^2 - 7x + 5 can be found using the formula: x = -b / (2a)
In this case, a = 3 and b = -7.
x = -(-7) / (2*3)
x = 7 / 6
Therefore, the x-value of the vertex is 7/6 or 1 1/6.