Find a such that f(x)= ax^2-3x+5 has a maximum value of 15.
I'm stuck because how am I suppose to complete the square? 3/2= 1.5? then square 1.5? So lost.
well, even without completing the square, you know the vertex is at x = -b/2a = 3/2a
So,
a(3/(2a))^2 - 3(3/(2a)) + 5 = 15
5 - 9/(4a) = 15
a = -9/40
But, in the spirit of completing the square,
ax^2-3x+5
= a(x^2 - 3/a) + 5
= a(x^2 - 3/a + (3/2a)^2) + 5 - a(3/2a)^2
= a(x - 3/2a)^2 + 5 - 9/4a
This is the vertex form
y = a(x-h)^2 + k
so we have
5 - 9/4a = 15
9/4a = -10
4a = -9/10
a = -9/40
the max is on the axis of symmetry
x = 3 / 2a
15 = a (3 / 2a)^2 - 3 (3 / 2a) + 5
10 = (9 / 4a) - (9 / 2a)
10 a = 9/4 - 9/2 = - 9/4
a = -9/40
To find the value of "a" that will result in a maximum value of 15 for the function f(x) = ax^2 - 3x + 5, you can use the process of completing the square. Here are the step-by-step instructions to help you solve this problem:
Step 1: Rewrite the function
Rewrite the function f(x) by grouping the terms without changing the overall expression:
f(x) = a(x^2 - (3/2)x) + 5
Step 2: Take half of the coefficient of x and square it
Take half of the coefficient of x (-3/2) and square it: (-3/2)/2 = (-3/4), then square it: (-3/4)^2 = 9/16.
Step 3: Add and subtract the value obtained in the previous step inside the parentheses:
f(x) = a(x^2 - (3/2)x + 9/16 - 9/16) + 5
Step 4: Factor and simplify the expression inside the parentheses:
f(x) = a[(x - (3/4))^2 - 9/16] + 5
Step 5: Distribute the "a" across the expression:
f(x) = a(x - (3/4))^2 - 9a/16 + 5
Step 6: The vertex form of a quadratic function is given by:
f(x) = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex. In this case, the vertex has a maximum value since the coefficient of x^2 is positive.
Step 7: Comparing the quadratic function in step 5 with the vertex form, you can see that the vertex is located at (3/4, -9a/16 + 5).
Step 8: Since the maximum value is given as 15, you know that the y-coordinate of the vertex should be 15.
Step 9: Set -9a/16 + 5 equal to 15 and solve for "a":
-9a/16 + 5 = 15
-9a/16 = 15 - 5
-9a/16 = 10
-9a = 160
a = -160/9
So the value of "a" that will result in a maximum value of 15 for the function f(x) = ax^2 - 3x + 5 is -160/9.
To find the value of a that results in the quadratic function f(x) = ax^2 - 3x + 5 having a maximum value of 15, we can use the concept of completing the square.
First, let's recall the general form of a quadratic function: f(x) = ax^2 + bx + c.
In this case, we have f(x) = ax^2 - 3x + 5.
To find the maximum value, we need to rewrite the quadratic function in vertex form, which is given by f(x) = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.
To complete the square, we'll start by focusing on the quadratic term, ax^2. We want to rewrite this term as a perfect square trinomial.
The coefficient of x^2 is a, so we divide the equation by a to isolate x^2: f(x)/a = x^2 - (3/a)x + 5/a.
Now, let's focus on the linear term, -(3/a)x. To complete the square, we take half of the coefficient of x, square it, and add it to both sides of the equation.
Taking half of -(3/a) gives us -3/2a. We square this value: (-3/2a)^2 = 9/4a^2.
Adding 9/4a^2 to both sides of the equation, we get: f(x)/a + 9/4a^2 = x^2 - (3/a)x + 5/a + 9/4a^2.
Now, we rewrite the right side of the equation as a perfect square trinomial: x^2 - (3/a)x + 5/a + 9/4a^2 = (x - (3/2a))^2 + (5/a + 9/4a^2).
At this point, we have rewritten f(x) in vertex form: f(x) = a(x - (3/2a))^2 + (5/a + 9/4a^2).
From the vertex form equation, we can determine that the vertex coordinates are (h, k), where h = 3/2a and k = 5/a + 9/4a^2.
To find the maximum value of 15, we need to set k equal to 15 and solve for a.
5/a + 9/4a^2 = 15.
To simplify the equation, we can multiply through by a to clear the denominators: 5 + (9/4)a = 15a.
Rearranging terms, we have: (9/4)a - 15a = -5.
Combining the like terms, we get: (9/4 - 60/4)a = -5.
Simplifying further, (9 - 60)a/4 = -5.
Therefore, (51a)/4 = -5.
To isolate a, we multiply both sides of the equation by 4/51: a = (-5)(4/51).
Finally, simplifying the expression to find the value of a: a = -20/51.
Therefore, for the quadratic function f(x) = ax^2 - 3x + 5 to have a maximum value of 15, the value of a must be -20/51.