Problem 43 Sketch the parabola. Label the v... [FREE SOLUTION]

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Elementary Algebra Within Reach

Ron Larson, Kimberly Nolting

$Math Studyset 91影视 Explanations$ Math

6 Edition

Chapter 10: Problem 43

Sketch the parabola. Label the vertex and any intercepts. $$y=-\left(x^{2}+4 x+4\right)$$

Short Answer

Expert verified

The vertex of the parabola is at $2, 0$ and it intersects the x-axis at $-2$. The parabola opens downwards.

Step by step solution

Find the vertex

The vertex of the parabola is given by $-h, 0$. In this case, comparing the equation $y=-\left(x^{2}+4 x+4\right)$ with $y = - (x + h)^2$, we find that $h = -2$. Therefore, the vertex of the parabola is at $-(-2), 0$ or $2, 0$.

Find the intercepts

To find the intercepts, we set $y = 0$. Therefore, we have $-\left(x^{2}+4 x+4\right) = 0$. Solving this we have $(x + 2)^2 = 0$. This gives $x = -2$ as the only intercept. So the parabola intercepts the x-axis at $-2$.

Sketch the parabola

Now, with the vertex and intercept known, we can sketch the parabola. The vertex is at the point $2, 0$. There is one x-intercept at $-2$. The parabola opens downwards as the coefficient of $x^2$ is negative. We can label these on our sketch.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex of a Parabola

The vertex of a parabola is a significant point where the parabola changes direction. In a quadratic equation of the form \[ y = ax^2 + bx + c \], the vertex can be found using specific formulas that depend on its standard or vertex form.

Standard form: Here, the vertex can be found using $ h = -\frac{b}{2a} $. The corresponding $ k $ value can be found by substituting $ h $ back into the equation to get $ k = y(h) $.
Vertex form: This is roughly of the form $ y = a(x-h)^2 + k $. The vertex ($ (h, k) $) is extracted directly from the equation.

In the original equation $ y = -(x^2 + 4x + 4) $, let's rewrite using vertex form to find $ h = -2 $. Thus, the vertex here is $ (-2, 0) $. This is usually a turning point for the parabola where it either reaches a maximum or minimum value.

X-Intercepts

Finding the x-intercepts allows us to determine where the parabola crosses the x-axis. An x-intercept can be found by setting $ y = 0 $ and solving the resulting equation for $ x $.

In our given equation $ -\left(x^2 + 4x + 4\right) = 0 $, simplify to $(x + 2)^2 = 0 $.
Solving this, we find $ x = -2 $. This indicates the parabola touches the x-axis at this point.

Unlike other quadratic equations which may have up to two x-intercepts, this equation only has one, indicating a special condition where the vertex lies directly on the x-axis.

Quadratic Equations

Quadratic equations, quintessential to parabolas, are polynomials of degree 2. They can describe the trajectory of a parabola in a coordinate plane.

The general form: $ y = ax^2 + bx + c $.
The standard form gives us essential information such as the direction and the shape of the parabola based on the coefficient of $ x^2 $.
A negative $ a $ means the parabola opens downwards.

Quadratic equations are solvable using various methods such as factoring, completing the square, and the quadratic formula. They represent real life scenarios like projectile motion or optics in physics.In our case,$ a $ being negative results in a downward-facing parabola, crucial to its sketching.

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91影视

Sketch the parabola. Label the vertex and any intercepts. $$y=-\left(x^{2}+4 x+4\right)$$

Short Answer

Step by step solution

Find the vertex

Find the intercepts

Sketch the parabola

Key Concepts

Vertex of a Parabola

X-Intercepts

Quadratic Equations

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