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Chapter 11: Problem 6

Identify the vertex of each parabola. $$ f(x)=-4 x^{2} $$

Short Answer

Expert verified
The vertex of the parabola \( f(x) = -4x^2 \) is at (0, 0).

Step by step solution

01

Understand the Standard Form of a Parabola

The standard form of a parabola is given by the equation: \[ f(x) = ax^2 + bx + c \] Here, the given equation is \( f(x) = -4x^2 \). Notice there are no \( bx \) or \( c \) terms, meaning \( b = 0 \) and \( c = 0 \).
02

Identify the Coefficients

From the given function \( f(x) = -4x^2 \), identify the coefficients: \[ a = -4, \, b = 0, \, c = 0 \]
03

Calculate the Vertex x-value

The x-coordinate of the vertex for a parabola in standard form is given by: \[ x = -\frac{b}{2a} \]Substitute the values of \( b \) and \( a \) into the equation: \[ x = -\frac{0}{2(-4)} = 0 \]
04

Calculate the Vertex y-value

Substitute the x-value of the vertex back into the function to find the y-coordinate. With \( x = 0 \), \[ f(0) = -4(0)^2 = 0 \]So, the y-coordinate of the vertex is 0.
05

Write the Vertex Coordinates

Combine the x and y values to write the vertex in coordinate form: \[ (0, 0) \]

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Key Concepts

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

Quadratic Function
A quadratic function is a type of polynomial function where the highest degree of the variable is squared. These functions are usually written in the form \[ f(x) = ax^2 + bx + c \]. In this formula:
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Most popular questions from this chapter

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (2 x-1)(8 x-4)=-1 $$

Solve each equation. (All solutions are nonreal complex numbers.) $$ (x+3)^{2}=-4 $$

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 2-2 x=3 x^{2} $$

Find the discriminant. Use it to determine whether the solutions for each equation are A. two rational numbers B. one rational number C. two irrational numbers D. two nonreal complex numbers. Tell whether the equation can be solved using the zero-factor property, or if the quadratic formula should be used instead. Do not actually solve. $$ 4 x^{2}-28 x+49=0 $$

Solve using the square root property. Simplify all radicals. $$ \left(x+\frac{1}{7}\right)^{2}=\frac{11}{49} $$
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