Consider the graph of the quadratic function y=ax^2+bx+c.
If it passes through two points P=(6,0) and Q=(14,0), and its vertex is on the line y=−16x, what is the value of a+b+c?
Since P and Q are x-intercepts, the x of the vertex must lie half-way between them
the x of the vertex is 10
since the vertex also lies on y = -16x
the vertex must be (10, -160)
You can now find the equation of the function,
comparing it to y = ax^2 + bx = c, the values of a, b, and c can be found.
May I know how to compare please? What are the steps? Thanks...
To find the value of a+b+c, we need to use the given information about the quadratic function. Let's break down the problem step by step:
Step 1: Understand the properties of a quadratic function.
A general quadratic function is represented by the equation y = ax^2 + bx + c. In this equation, a, b, and c are constants that determine the shape and position of the graph.
Step 2: Use the given points to form equations.
We have two points on the graph: P(6, 0) and Q(14, 0). Since these points lie on the graph of the quadratic function, we can substitute the x and y values into the equation to create two equations.
For point P(6, 0): 0 = a(6)^2 + b(6) + c
For point Q(14, 0): 0 = a(14)^2 + b(14) + c
These two equations will help us solve for the values of a, b, and c.
Step 3: Understand the vertex of the quadratic function.
The vertex of a quadratic function is the highest or lowest point on its graph. We are given that the vertex lies on the line y = -16x.
Step 4: Determine the x-coordinate of the vertex.
Since the vertex lies on the line y = -16x, we can substitute this equation into the quadratic function equation to find the x-coordinate of the vertex.
y = ax^2 + bx + c
-16x = ax^2 + bx + c
Now, we need to find the x-coordinate of the vertex, which can be found using the formula x = -b/2a. In this case, we have a = 1 and b = -16. Plugging these values into the formula, we get:
x = -(-16)/2(1) = 8
The x-coordinate of the vertex is 8.
Step 5: Find the y-coordinate of the vertex.
Since the vertex lies on the line y = -16x, we can substitute the x-coordinate of the vertex (x = 8) into this equation to find the y-coordinate.
y = -16(8) = -128
The y-coordinate of the vertex is -128.
Step 6: Formulate an equation using vertex form.
The vertex form of a quadratic function is y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex. We have the coordinates of the vertex: (8, -128). Substituting these values in, we get:
y = a(x - 8)^2 - 128
Step 7: Solve for a.
We can use one of the given points (P or Q) to solve for a. Let's use point P(6, 0). Substituting this point into the quadratic function equation, we get:
0 = a(6 - 8)^2 - 128
0 = a(-2)^2 - 128
0 = 4a - 128
4a = 128
a = 32
Step 8: Find b and c.
We can now substitute the values of a and the coordinates of point P (6, 0) into the quadratic function equation to solve for b and c.
0 = 32(6 - 8)^2 - 128
0 = 32(-2)^2 - 128
0 = 32(4) - 128
0 = 128 - 128
0 = 0
Substituting the coordinates of point P into the quadratic function equation, we get:
0 = 32(6)^2 + b(6) + c
0 = 32(36) + 6b + c
0 = 1152 + 6b + c
Since the y-coordinate of point P is 0, we can simplify this equation to:
c = -1152 - 6b
Now we have two equations involving b and c. Let's combine them to solve for b.
0 = -1152 - 6b
6b = -1152
b = -192
Now that we have the values of a, b, and c, we can find their sum:
a + b + c = 32 + (-192) + c
Given that c = -1152 - 6b, substitute this value in:
a + b + c = 32 + (-192) + (-1152 - 6b)
Now, substitute b = -192:
a - 192 + (-1152 - 6(-192))
a - 192 - 1152 + 1152
a - 192 - 1152
a - 1344
Therefore, the value of a + b + c is a - 1344.