Question

Квадратная функция y=x²+6x+5 решить за дескрименантом и построить график

Answer 1

файл

--------------------------------------

Answer 2

Quadratic Function: $y = x^2 + 6x + 5$

To solve the quadratic function $y = x^2 + 6x + 5$ using the discriminant and to plot its graph, we can follow these steps:

1. Find the discriminant ($D$) of the quadratic function using the formula $D = b^2 - 4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic function.

For the given quadratic function $y = x^2 + 6x + 5$, the coefficients are: - $a = 1$ - $b = 6$ - $c = 5$

Substituting these values into the formula, we get: $D = (6)^2 - 4(1)(5) = 36 - 20 = 16$

The discriminant is $D = 16$.

2. Determine the nature of the roots based on the value of the discriminant: - If $D > 0$, the quadratic function has two distinct real roots. - If $D = 0$, the quadratic function has one real root (a perfect square). - If $D < 0$, the quadratic function has no real roots (complex roots).

In this case, since $D = 16 > 0$, the quadratic function has two distinct real roots.

3. Find the roots of the quadratic function using the quadratic formula: The quadratic formula is given by: $x = \frac{-b \pm \sqrt{D}}{2a}$

Substituting the values of $a$, $b$, and $D$ into the formula, we get: $x = \frac{-6 \pm \sqrt{16}}{2(1)}$

Simplifying further, we have: $x = \frac{-6 \pm 4}{2}$

This gives us two solutions: - $x_1 = \frac{-6 + 4}{2} = -1$ - $x_2 = \frac{-6 - 4}{2} = -5$

Therefore, the roots of the quadratic function are $x = -1$ and $x = -5$.

4. Plotting the graph of the quadratic function: To plot the graph, we can start by finding the vertex of the parabola. The vertex of a quadratic function in the form $y = ax^2 + bx + c$ is given by the coordinates $\left(-\frac{b}{2a}, -\frac{D}{4a}\right)$.

For the given quadratic function $y = x^2 + 6x + 5$, the coefficients are: - $a = 1$ - $b = 6$ - $c = 5$

Substituting these values into the formula, we get: Vertex $= \left(-\frac{6}{2(1)}, -\frac{16}{4(1)}\right) = (-3, -4)$

The vertex of the parabola is at the point $(-3, -4)$.

We can also find the $y$-intercept by substituting $x = 0$ into the quadratic function: $y = (0)^2 + 6(0) + 5 = 5$

Therefore, the $y$-intercept is at the point $(0, 5)$.

With the vertex and $y$-intercept, we can plot the graph of the quadratic function $y = x^2 + 6x + 5$.


The graph of the quadratic function $y = x^2 + 6x + 5$ is a parabola that opens upwards. It intersects the $x$-axis at $x = -5$ and $x = -1$, and it intersects the $y$-axis at $y = 5$. The vertex of the parabola is at the point $(-3, -4)$.

Please note that the graph is a visual representation and may not be to scale.