Parabola y=(x-p)²+k Practice Problems & Vertex Form

To solve this problem, let us first note that the labeled point is $(0, -4)$, which suggests the parabola touches or intersects the y-axis at this point. Without more information indicating horizontal translation, it is reasonable to assume this is the vertex of the parabola, pointing down a simple transformation from $y=x^2$ to $y=x^2-4$.

Given the simplicity and symmetry (likely no $x$ coefficient subtracted or added), this directly translates to a parabola form with only a vertical shift downward.

Therefore, the algebraic representation of the given parabolic drawing is $y = x^2 - 4$.

The correct choice corresponding to this is $y = x^2 - 4$.

To solve this problem, follow these steps:

• Identify the point related to the parabola, which is given as $(5, 4)$.
• This point is likely the vertex of the parabola. The vertex form equation is $y = (x-h)^2 + k$.
• Substitute the vertex coordinates $(h, k) = (5, 4)$ into the vertex form.

Using these steps, substitute $h = 5$ and $k = 4$ into the vertex form:

$y = (x - 5)^2 + 4$

This matches the given point and reflects the parabola intersecting or having its vertex at (5, 4).

Therefore, the algebraic representation of the drawing is $y = (x-5)^2 + 4$.

To determine the algebraic representation, we use the vertex form of a parabola, which is $y = (x-h)^2 + k$. Here, the vertex is placed at $(-2, 7)$, thus plug these values into our equation: $h = -2$ and $k = 7$.

Consequently, the equation of the parabola becomes:

$y = (x + 2)^2 + 7$

This representation correctly describes a parabola that passes through the vertex at $(-2, 7)$ and opens upwards, as indicated by the absence of a negative sign or alternate coefficient in front of the square term.

Therefore, the correct choice corresponding to this problem formulation is:

$y = (x + 2)^2 + 7$

To solve this problem, the following steps are necessary:

We begin with the original function:

• $y = -x^2$

First, we apply the horizontal shift of 3 units to the left. Moving a graph left involves adding a number to $x$ in the equation. Hence, replace $x$ with $(x + 3)$. This manipulatively affects the original function:

$y = -(x + 3)^2$

Next, we apply the vertical shift of 4 units upward. This involves adding 4 to the function:

$y = -(x + 3)^2 + 4$

Therefore, the equation representing the parabola moved 3 spaces to the left and 4 spaces up is:

$y = -(x + 3)^2 + 4$

Verification against the choices confirms that the correct answer is choice (1):

• $y = -(x + 3)^2 + 4$

This is indeed the equation that results after applying the given transformations to the original function $y = -x^2$.

To solve this problem, we'll start by understanding the transformations required:

• The original function is $y = x^2$.
• We need to move this function 2 spaces to the right and 5 spaces upwards.

Step 1: Apply the horizontal shift 2 units to the right.
To shift a function horizontally, replace $x$ with $x - h$, where $h$ is the shift to the right. Thus, we replace $x$ with $x - 2$ to get:

$y = (x - 2)^2$.

Step 2: Apply the vertical shift 5 units upwards.
To shift a function vertically, add $k$ to the function, where $k$ is the number of units to shift up. Thus:

$y = (x - 2)^2 + 5$.

Combining these transformations, the equation of the transformed function is:

$y = (x - 2)^2 + 5$.

This matches the choice labeled as 3. Thus, the correct equation after translating the parabola 2 spaces to the right and 5 spaces upwards is:

$y = (x - 2)^2 + 5$.