Increasing and Decreasing Domain of a Parabola: Standard representation

The function given is $y = 3x^2 - 6x + 4$.

First, let's find the derivative of the function, which will help us determine the intervals of increase.

The derivative is given by $f'(x) = \frac{d}{dx}(3x^2 - 6x + 4) = 6x - 6$.

Next, find where the derivative is zero to locate critical points. Solve $6x - 6 = 0$ to get:

$6x = 6$
$x = 1$

The critical point is $x = 1$. This is where the function changes from decreasing to increasing since quadratic functions have one axis of symmetry and $a > 0$: indicating a parabola opening upwards.

To determine the interval of increase, analyze the sign of $f'(x)$:

• For $x > 1$, $6x - 6 > 0$ which implies $f'(x) > 0$, so the function is increasing.
• For $x < 1$, $6x - 6 < 0$ which implies $f'(x) < 0$, so the function is decreasing.

Thus, the domain of increase for the function is when $x > 1$.

The correct answer is therefore $x > 1$.

To determine the domain over which the quadratic function $y = -x^2 + 2x + 35$ is decreasing, we proceed by identifying the vertex of the parabola.

Given the form $y = ax^2 + bx + c$, we have $a = -1$, $b = 2$, and $c = 35$. The x-coordinate of the vertex can be found using the formula:

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

Substituting $b = 2$ and $a = -1$ into the formula, we calculate:

$x = -\frac{2}{2 \times (-1)} = -\frac{2}{-2} = 1$

The vertex of the parabola occurs at $x = 1$. Since the function is a downward-opening parabola (as indicated by the negative coefficient of $-x^2$), the function decreases for all $x$ values greater than the x-coordinate of the vertex.

Therefore, the domain of decrease for the function is $x > 1$.

This matches the answer choice:

$x > 1$

To find the domain of increase for the function $y = -x^2 + 2x + 35$, let's determine the vertex first.

• Step 1: Identify coefficients in the quadratic equation. Here, $a = -1$, $b = 2$, and $c = 35$.
• Step 2: Use the vertex formula $x = -\frac{b}{2a}$ to find the x-coordinate of the vertex.

Plug in the values for $b$ and $a$:

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

The x-coordinate of the vertex is $x = 1$.

Since the coefficient $a$ is negative, this means the parabola opens downwards. A parabola opening downward will increase until it reaches the vertex, then start decreasing.

Therefore, the domain on which the function is increasing is $x < 1$.

Therefore, the solution to the problem is $x < 1$.