Given the expression of the quadratic function
Finding the symmetry point of the function
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Given the expression of the quadratic function
Finding the symmetry point of the function
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: The given function is , where and .
Step 2: The axis of symmetry for a quadratic function in the form is given by . With , this simplifies to .
Step 3: To find the vertex, calculate the function's value at , using .
Plugging in , we find:
.
Thus, the vertex, and hence the symmetry point of the function, is .
Therefore, the solution to the problem is .
Given the expression of the quadratic function
Finding the symmetry point of the function
\( f(x)=-5x^2+10 \)
The coefficient 2 affects the parabola's width, not its vertex location. Since there's no x term (b = 0), the vertex lies on the y-axis at .
The symmetry point is the same as the vertex! It's where the parabola is perfectly balanced - you can fold the graph along this point and both sides match exactly.
Think: "negative b over 2a" gives you the x-coordinate. Then substitute that x-value back into the function to find the y-coordinate of the vertex.
Then b = 4, so . The vertex would be at x = -1, not x = 0. The b term shifts the vertex sideways!
Only when a > 0 (parabola opens upward). Since a = 2 is positive here, (0,0) is indeed the lowest point. If a were negative, it would be the highest point.
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