How do you find the domain and range of #y = −3x^2 − 3x + 4#?

Question

Ethan Adkins · Accepted Answer

This is a quadratic funtion and the domain is all the Real #x#. This means that you can give every value of #x# in the real domain and your function devolves a value of #y#. The graph of a quadratic is called a PARABOLA, or broadly speaking, a U-shaped curve. The range is a little bit tricky. The position of the lowes (or highest) point gives us the possibility to "see" the range!
Your quadratic has #-3# in front of #x^2# so it is a "sad" parábola, a inverted U shaped curve
The highest point is called the Vertex and is given (the coordinates) as (if you have your quadratic in the general form:
#ax^2+bx+c=0#):
#x_v=-b/(2a)# 
#y_v=-Delta/(4a)#
with #Delta=b^2-4ac# in your case:
#x_v=-(-3)/(2*-3)=-1/2#
#y_v=[9-4(-3*4)]/(4*-3)=4.75# So your range is all the #y# less or equals to #4.75#. A graph representing this is graph{-3x^2-3x+4 [-10.62, 7.16, 1.22, 10.11]}.

Eva Abramson · Answer

undefined To find the domain and range of the function y = -3x^2 - 3x + 4, you first determine the domain by identifying all possible real values of x. Since it's a quadratic function, the domain is all real numbers. 
To find the range, analyze the behavior of the function. Since the coefficient of x^2 is negative, the parabola opens downward, so the maximum value of the function occurs at the vertex. The vertex of the parabola can be found using the formula x = -b/(2a), where a = -3 and b = -3. Substitute x = -3/(2*(-3)) = 1/2 into the equation to find the corresponding y-value. Then, the range is all real numbers less than or equal to this y-value. So, the range is y ≤ 19/4.

Eva Abramson · Answer

undefined To find the domain and range of the function $y = -3x^2 - 3x + 4$:

**Domain:**
The domain of a quadratic function is all real numbers, because you can substitute any real number for $x$ into the equation without encountering any mathematical restrictions such as division by zero or taking the square root of a negative number. Thus, the domain of $y = -3x^2 - 3x + 4$ is $\mathbb{R}$, or in interval notation, $(-∞, ∞)$.

**Range:**
The range of a quadratic function depends on the direction of the parabola (opening upwards or downwards) and its vertex. Since the coefficient of $x^2$ is negative ($-3$), the parabola opens downwards. This means the function's maximum value occurs at the vertex of the parabola.

To find the vertex, use the formula $x = \frac{-b}{2a}$ for the x-coordinate of the vertex, where $a = -3$ and $b = -3$. Plugging in these values gives:

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

To find the y-coordinate of the vertex (which is the maximum value of the function since the parabola opens downwards), substitute $x = -\frac{1}{2}$ back into the original equation:

$$y = -3\left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 4$$

$$y = -3\left(\frac{1}{4}\right) + \frac{3}{2} + 4$$

$$y = -\frac{3}{4} + \frac{3}{2} + 4$$

$$y = -\frac{3}{4} + \frac{6}{4} + \frac{16}{4}$$

$$y = \frac{19}{4}$$

So, the vertex is at $\left(-\frac{1}{2}, \frac{19}{4}\right)$, meaning the highest point on the graph is $\frac{19}{4}$. Therefore, the range of the function is all real numbers less than or equal to $\frac{19}{4}$, or in interval notation, $(-∞, \frac{19}{4}]$.