Which parabola is the translation of the graph of the function
and is positive in all areas except?
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Which parabola is the translation of the graph of the function
and is positive in all areas except?
To solve this problem, we'll follow these steps:
Now, let's work through each step:
Step 1: We are given the function . The new parabola must be zero at and positive everywhere else. This means the vertex of the parabola is at .
Step 2: To move the vertex of from to , we need a horizontal shift to the left by 3 units. The translation is represented by replacing with in the function:
This new equation reflects a parabola that opens upwards and has its vertex at , which means it is zero only at and positive everywhere else.
Therefore, the solution to the problem is .
Find the intersection of the function
\( y=(x+4)^2 \)
With the Y
This confuses many students! Think of it this way: (x + 3)² = 0 when x = -3. The parabola reaches its minimum (vertex) where the expression inside equals zero. So x + 3 = 0 means x = -3, placing the vertex at x = -3.
Use the "opposite rule": The sign inside the parentheses is opposite to the shift direction. (x + 3) shifts LEFT 3 units, (x - 3) shifts RIGHT 3 units. Always solve for when the expression equals zero!
This means the parabola touches the x-axis at exactly one point: x = -3. Everywhere else, the y-values are above the x-axis (positive). The vertex must be at (-3, 0) with the parabola opening upward.
No! While y = -(x + 3)² is zero at x = -3, it's negative everywhere else (opens downward). We need the parabola to be positive except at x = -3, so it must open upward.
Test three points: x = -3 (should give y = 0), x = -2 (should be positive), and x = -4 (should be positive). For y = (x + 3)²: at x = -2, y = 1 ✓; at x = -4, y = 1 ✓
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