Math Problem Statement

Sketch the graph of the quadratic function. (Select Update Graph to see your response plotted on the screen. Select the Submit button to grade your response.) f(x) = x2 + x − 7 Indicate the coordinates of the vertex, the y-intercept, and the x-intercepts (if any). (If an answer does not exist, enter DNE. If there are multiple x-intercepts, enter your answers as a comma-separated list.) vertex (x, y)

y-intercept y = x-intercepts x = Since the graph is symmetric around the vertical line through the vertex, reflecting the y-intercept across the line of symmetry, we find the additional point (x, y) =

on the graph. Determine the left-hand and right-hand behavior of the graph. (If an answer does not exist, enter DNE. If you need to use or –, enter INFINITY or –INFINITY, respectively.) As the graph goes to the left without bound, the values of y approach . As the graph goes to the right without bound, the values of y approach .

Solution

Let's break down and analyze the quadratic function f(x)=x2+x−7f(x) = x^2 + x - 7 step by step, starting with the key features: vertex, intercepts, symmetry, and behavior of the graph.

1. Finding the Vertex

The vertex of a parabola f(x)=ax2+bx+cf(x) = ax^2 + bx + c can be found using the formula for the x-coordinate of the vertex:

xvertex=−b2ax_{\text{vertex}} = -\frac{b}{2a}

For the given quadratic equation f(x)=x2+x−7f(x) = x^2 + x - 7, the coefficients are a=1a = 1, b=1b = 1, and c=−7c = -7.

xvertex=−12(1)=−12x_{\text{vertex}} = -\frac{1}{2(1)} = -\frac{1}{2}

Now, substitute x=−12x = -\frac{1}{2} back into the equation to find the y-coordinate of the vertex:

= \frac{1}{4} - \frac{1}{2} - 7 = \frac{1}{4} - \frac{2}{4} - \frac{28}{4} = \frac{-29}{4}$$ Thus, the coordinates of the vertex are: $$\text{Vertex: } \left( -\frac{1}{2}, -\frac{29}{4} \right)$$ ### 2. **Finding the y-intercept** The y-intercept occurs when $$x = 0$$. Substituting $$x = 0$$ into the equation: $$y_{\text{intercept}} = f(0) = 0^2 + 0 - 7 = -7$$ Thus, the y-intercept is: $$y = -7$$ ### 3. **Finding the x-intercepts** The x-intercepts occur when $$f(x) = 0$$. To find the x-intercepts, solve the equation $$x^2 + x - 7 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Substitute the values $$a = 1$$, $$b = 1$$, and $$c = -7$$: $$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-7)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 28}}{2} = \frac{-1 \pm \sqrt{29}}{2}$$ Thus, the two x-intercepts are: $$x = \frac{-1 + \sqrt{29}}{2}, \quad x = \frac{-1 - \sqrt{29}}{2}$$ These are the approximate values: $$x \approx 2.19, \quad x \approx -3.19$$ So, the x-intercepts are approximately: $$x \approx 2.19, -3.19$$ ### 4. **Reflecting the y-intercept Across the Line of Symmetry** The graph is symmetric about the vertical line through the vertex $$x = -\frac{1}{2}$$. The y-intercept is at $$(0, -7)$$. To find the reflected point, calculate how far 0 is from $$x = -\frac{1}{2}$$, and reflect that distance on the other side: $$\text{Distance from } 0 \text{ to } -\frac{1}{2} = \frac{1}{2}$$ So, reflect to the left of the vertex by 0.5 units: $$x = -1$$ Thus, the reflected point is: $$(x, y) = (-1, -7)$$ ### 5. **Left-hand and Right-hand Behavior** Since the coefficient of $$x^2$$ is positive ($$a = 1$$), the parabola opens upward. The left-hand and right-hand behavior of the graph are: - As $$x \to -\infty$$, $$f(x) \to \infty$$ - As $$x \to \infty$$, $$f(x) \to \infty$$ ### Final Summary: - Vertex: $$\left( -\frac{1}{2}, -\frac{29}{4} \right)$$ - y-intercept: $$y = -7$$ - x-intercepts: $$x \approx 2.19, -3.19$$ - Reflected point: $$(-1, -7)$$ - Left-hand behavior: As $$x \to -\infty$$, $$y \to \infty$$ - Right-hand behavior: As $$x \to \infty$$, $$y \to \infty$$ Would you like more details on any of these steps or help plotting the graph? Here are 5 questions to deepen your understanding: 1. How can the quadratic formula be applied to other types of quadratics? 2. Why does the parabola open upward in this case? 3. What role does the discriminant play in determining the number of x-intercepts? 4. How does the vertex form of a quadratic function help in graphing? 5. Can you derive the axis of symmetry from the general quadratic equation? **Tip:** When solving quadratic equations, always check the discriminant ($$b^2 - 4ac$$) first to determine whether the equation has real, imaginary, or repeated roots.

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Math Problem Analysis

Mathematical Concepts

Quadratic Functions
Graphing Parabolas
Intercepts
Symmetry

Formulas

Vertex formula: x_vertex = -b / (2a)
Quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula Theorem
Symmetry of Parabolas

Suitable Grade Level

Grades 9-12

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