Parabola of the Form y=(x-p)²: True / false

To solve this problem, we need to determine whether substituting $x = 0$ gives us the points of intersection with the x-axis for the parabola $y = (x - p)^2$.

To find the x-intercepts of any function, we set $y = 0$ because the x-intercepts occur where the curve meets the x-axis, which implies a zero output or function value:

• Start with the equation: $y = (x - p)^2$.
• Set $y = 0$ for the x-intercept: $(x - p)^2 = 0$.
• Solving this gives $(x - p) = 0$, which simplifies to $x = p$.
• Thus, the x-intercept happens at the point where $x = p$, not where $x = 0$.

Therefore, substituting $x = 0$ does not provide the x-intercepts. Instead, it provides the y-intercept. Therefore, the statement given in the problem is False.

Thus, the solution to the problem is False.

To determine if the given statement is true, consider the equation of the parabola $y = (x - p)^2$. The y-intercept occurs where the parabola crosses the y-axis, which is at $x = 0$.

Step 1: Substitute $x = 0$ into the equation:

$y = (0 - p)^2 = p^2$

Step 2: Calculate the y-intercept:

The y-intercept of the parabola is $y = p^2$.

Conclusion: The statement "To find the y-axis intercept, you substitute $x = 0$ into the equation and solve for $y$" is indeed True, as applying this method correctly determined the y-intercept for the given form of a parabola. Therefore, the answer to the problem is True.

To solve this problem, we need to determine whether knowing the vertex of a function $y = (x-p)^2$ allows us to conclude if the function's domain is ascending or descending.

Consider the vertex form of the given parabola: $y = (x-p)^2$.

• The parabolic function opens upward since the coefficient of $(x-p)^2$ is positive (1).
• The vertex of the parabola is $(p, 0)$.
• A parabola's opening direction is determined by the sign of the coefficient of the squared term. By understanding this, we ascertain that the function decreases towards the vertex (left side of vertex) and increases away (right side of vertex).
• Thus, the left side of $p$, as $x$ approaches the vertex, is descending, and to the right is ascending, confirming the vertex as a divider of these behaviors.

Overall, knowing the vertex allows us to describe the behavior of the function's graph as ascending or descending at and away from the vertex point, verifying the statement.

The conclusion is that the statement is True.

The given functions are $y = x^2$ and $y = 2x^2$. Both functions are written in the form $y = ax^2$, where the vertex for such parabolas is determined by setting $x = 0$.

Let's analyze each function step-by-step:

• Function $y = x^2$:
  This function represents a parabola that opens upwards with its vertex located at $(0, 0)$.
• Function $y = 2x^2$:
  Similarly, this function is also a parabola that opens upwards. The coefficient $2$ affects the "narrowness" or "width" of the parabola but does not shift the vertex, so the vertex remains at $(0, 0)$.

Both functions $y = x^2$ and $y = 2x^2$ share the same vertex at the point $(0, 0)$. Therefore, the statement that the vertex of the function $y=x^2$ is the same vertex of the function $y=2x^2$ is True.

Thus, the correct answer to the problem is True.

To solve this problem, we'll verify if $y = (x-4)^2$ is obtained by shifting $y = x^2$ four units to the right:

• Step 1: Identify transformation properties.
  The equation $y = (x-h)^2$ indicates a horizontal shift of the parent function $y = x^2$.
• Step 2: Determine the shift value from the expression.
  In $y = (x-4)^2$, $h = 4$. This implies a shift 4 units to the right because $h = 4$ is positive.
• Step 3: Verify the shift.
  If we replaced $x$ with $x-4$, this represents taking the graph of $y = x^2$ and shifting it rightwards by 4 units on the x-axis.

Therefore, we conclude that the function $y = (x-4)^2$ does represent a displacement of $y = x^2$ to the right by 4 units. Hence, the correct answer to the problem is True.

To solve this problem, we'll determine how the transformation $y = (x+4)^2$ affects the graph of $y = x^2$.

The function $y = x^2$ is a standard parabola centered at the origin.

In $y = (x+4)^2$, the positive number inside the parentheses indicates a transformation that moves the entire graph horizontally.

• Specifically, the expression $(x + 4)$ means we take the original $x$ value and add 4 to it before squaring, effectively shifting the graph.

According to properties of horizontal translations, when you add a positive number to $x$ inside the function—here, the +4 in $(x+4)$—the graph of the function $y = x^2$ shifts 4 units to the left along the x-axis.

Therefore, the transformation described by $y = (x+4)^2$ is indeed the graph of $y = x^2$ moved 4 spaces to the left.

Thus, the correct answer to this problem is Yes.