Plotting Functions with Parameters: Linking function properties to its representation

To determine if the function $y = (x-2)(x+1)$ has a minimum or maximum point, we start by converting it from product form to standard form:

$y = (x-2)(x+1)$

Expanding the expression:

$y = x^2 + x - 2x - 2$

Simplify:

$y = x^2 - x - 2$

In standard form, $y = x^2 - x - 2$, the coefficient of $x^2$, which is $a = 1$, is positive. A positive $a$ indicates the parabola opens upwards.

Since the parabola opens upwards, it has a minimal point (vertex) as its lowest point.

Therefore, the parabola $y = (x-2)(x+1)$ has a minimal point.

The quadratic function is given by $y = -x^2 + 3x + 9$.

In the general quadratic form $y = ax^2 + bx + c$, the coefficient $a$ determines the parabola's orientation:

• If $a > 0$: The parabola opens upwards, indicating a minimum point.
• If $a < 0$: The parabola opens downwards, indicating a maximum point.

For this function, $a = -1$. Since $a < 0$, the parabola opens downwards.

Therefore, the function has a maximum point.

Thus, the correct answer is the function has a highest point.

Therefore, the solution to the problem is choice 2: Highest point.

To determine whether the quadratic function $y = x^2 + 2x$ has a minimum or maximum point, we need to examine its structure and calculate the vertex.

Step 1: Identify the structure of the quadratic function.

The given function is $y = x^2 + 2x$, which is a standard form quadratic function $y = ax^2 + bx + c$ where $a = 1$, $b = 2$, and $c = 0$.

Step 2: Calculate the vertex.

The vertex of a quadratic function is given by $x = -\frac{b}{2a}$. Substituting the values of $a$ and $b$ into this formula gives:

$x = -\frac{2}{2 \times 1} = -1$.

Substitute $x = -1$ back into the original equation to find the y-coordinate of the vertex:

$y = (-1)^2 + 2(-1) = 1 - 2 = -1$.

Therefore, the vertex is at the point $(-1, -1)$.

Step 3: Determine if the vertex is a minimum or maximum.

Since the coefficient $a = 1$ is positive, the parabola opens upwards. This means that the vertex represents the lowest point on the graph, which is a minimum point.

Therefore, the solution to this problem is that the parabola has a minimal point.

To determine whether there is a minimum or maximum point in the quadratic function $y = -2x(x+1)$, let's follow these steps:

• Simplify the given function. Distribute the negative sign to expand the polynomial:

$y = -2x(x+1)$

$y = -2x^2 - 2x$

• Identify the coefficients of the quadratic equation in the form $ax^2 + bx + c$:
• $a = -2$, $b = -2$, and $c = 0$.
• Since the coefficient of $x^2$, $a$, is negative, the parabola opens downwards.

When a parabola opens downwards, it has a maximum point, as it reaches a peak value (the highest point) before descending.

Therefore, the quadratic function $y = -2x(x+1)$ has a maximum point.

Hence, the correct answer is that the parable has a highest point.

To solve this problem, we need to determine whether the quadratic function $y = 6x - x^2$ has a minimum or maximum point.

Firstly, let's identify the general form of the quadratic equation, which is $y = ax^2 + bx + c$. For our function, we have:

• $a = -1$
• $b = 6$
• $c = 0$

The coefficient $a = -1$ is negative, indicating that the parabola opens downwards. A downward-opening parabola means that the function has a maximum point.

Next, we calculate the x-coordinate of the vertex, which gives the maximum point:

The formula for the x-coordinate of the vertex of a quadratic function $ax^2 + bx + c$ is:

$x = -\frac{b}{2a}$

Substituting the values of $a$ and $b$ into the vertex formula, we get:

$x = -\frac{6}{2 \times -1} = -\frac{6}{-2} = 3$

Thus, the x-coordinate of the vertex is $x = 3$. We can substitute this back into the original equation to find the y-coordinate:

$y = 6(3) - (3)^2 = 18 - 9 = 9$

Therefore, the vertex of the parabola is at $(3, 9)$, and since the parabola opens downwards, this point represents the maximum point of the function.

We conclude that the quadratic function $y = 6x - x^2$ reaches a highest point.

To solve this problem, follow these steps:

• Step 1: Express the quadratic function in standard form. The given function is $y = 2x(x + 3)$, which can be expanded to $y = 2x^2 + 6x$.
• Step 2: Identify the coefficient of $x^2$. In this case, $a = 2$.
• Step 3: Determine the direction of the parabola. Since $a = 2$ is positive, the parabola opens upwards.
• Step 4: Conclude that the vertex represents the lowest point, which is the minimum point.

Given that the coefficient $a$ is positive, the parabola opens upwards, indicating that the vertex is a minimum point.

Therefore, the solution to this problem is minimal point.

To solve this problem, let's analyze the function step-by-step:

Step 1: Expand the given quadratic function.
The function provided is $y = -(x+1)(x-1)$. We rewrite it by expanding:

$y = -(x^2 - 1) = -x^2 + 1$.

Step 2: Determine the direction of the parabola.
The standard form $y = ax^2 + bx + c$ indicates that the parabola opens upwards if $a > 0$ and downwards if $a < 0$. Here, the value of $a$ is $-1$, which means the parabola opens downwards.

Step 3: Identify the vertex type.
Since the parabola opens downwards, the vertex represents the highest point on the graph, which is a maximum.

Therefore, the parabola has a maximum point. The correct choice is:

$$Choice 2: Highest point$$

Thus, we conclude that the given quadratic function has a maximum point.

To solve this problem, follow these steps:

• Step 1: Expand the quadratic function.
• Step 2: Determine the nature based on the leading coefficient.

Step 1: Expand the quadratic function:

The given function is $y = (x+1)(-x-1)$.

Expanding this, we have:

$y = (x+1)(-x-1) = x(-x-1) + 1(-x-1) = -x^2 - x - x - 1 = -x^2 - 2x - 1$

Step 2: Determine the nature using the leading coefficient:

The quadratic function is $y = -x^2 - 2x - 1$.

Here, the leading coefficient $a = -1$. Since the leading coefficient is negative, the parabola opens downwards.

Therefore, the quadratic function has a highest point, or a maximum.

Thus, the solution to the problem is that the quadratic function has a highest point.