Question

Y=x^2+4x+3 график функций По графикунайти: а) нули функции б) множество значений функции Е

Answer 1

График во вложении

а) х=0 при у=3,  у=0 при х=-1;-3

б) Множество значений функции от - бесконечности до +бесконечности

Answer 2

To find the zeros of the function and the range (set of values) of the function, we'll first analyze the given function:

$y = x^2 + 4x + 3$

a) Zeros of the function (x-intercepts): The zeros of the function are the values of 'x' for which the function equals zero ($y = 0$). In other words, we need to find the x-values where the graph of the function intersects the x-axis.

To find the zeros, set $y = 0$ and solve for 'x':

$0 = x^2 + 4x + 3$

We can factor the quadratic equation or use the quadratic formula to find the zeros. In this case, let's use the quadratic formula:

The quadratic formula states that for an equation $ax^2 + bx + c = 0$, the solutions for 'x' are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our equation, $a = 1$, $b = 4$, and $c = 3$.

$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$ $x = \frac{-4 \pm \sqrt{16 - 12}}{2}$ $x = \frac{-4 \pm \sqrt{4}}{2}$ $x = \frac{-4 \pm 2}{2}$

Now, we have two possible values for 'x':

1. $x = \frac{-4 + 2}{2} = \frac{-2}{2} = -1$
2. $x = \frac{-4 - 2}{2} = \frac{-6}{2} = -3$

So, the zeros of the function are $x = -1$ and $x = -3$.

b) Range of the function (set of values): To find the range of the function, we need to determine the set of all possible 'y' values that the function can take as 'x' varies over the real numbers.

Since the coefficient of $x^2$ is positive (1 in this case), the parabola opens upwards, and its vertex represents the minimum value. As there are no restrictions on 'x', the minimum value of the function will be negative infinity.

To find the range, we can find the value of the function at the vertex and determine the set of all real numbers greater than or equal to that value.

The x-coordinate of the vertex can be found using $x = \frac{-b}{2a}$:

$x = \frac{-4}{2 \cdot 1} = -2$

Now, to find the corresponding 'y' value at the vertex, substitute $x = -2$ into the original function:

$y = (-2)^2 + 4(-2) + 3$ $y = 4 - 8 + 3$ $y = -1$

So, the vertex is at (-2, -1), and the minimum value of the function is -1.

Thus, the range of the function is all real numbers greater than or equal to -1:

$E = \{y \mid y \geq -1\}$

In interval notation, the range is $(-1, +\infty)$.