/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 145 Find the vertex of the graph of ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the vertex of the graph of \(f(x)=-x^{2}+3 x-10\) and determine the interval on which the function is increasing.

Short Answer

Expert verified
The vertex is \(\left(\frac{3}{2}, -\frac{31}{4}\right)\). The function is increasing on \(( -\infty, \frac{3}{2})\).

Step by step solution

01

Identify the coefficients

For the quadratic function in the form of \(ax^2 + bx + c\), identify the coefficients \(a = -1\), \(b = 3\), and \(c = -10\).
02

Determine the vertex (h, k)

Use the vertex formula \(h = -\frac{b}{2a}\). Substitute \(a = -1\) and \(b = 3\) to find \(h\). Thus, \[h = -\frac{3}{2(-1)} = \frac{3}{2} \].
03

Calculate k

Substitute \(h = \frac{3}{2}\) back into the function \(f(x) = -x^2 + 3x - 10\) to find \(k\). \[k = f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) - 10 \] \[\Rightarrow k = -\frac{9}{4} + \frac{9}{2} - 10 = -\frac{9}{4} + \frac{18}{4} - \frac{40}{4} = -\frac{31}{4} \]
04

Vertex coordinates

The vertex is therefore at \(\left(\frac{3}{2}, -\frac{31}{4}\right)\).
05

Determine the interval of increase

For a quadratic function \(ax^2 + bx + c\) with \(a < 0\), it opens downward. The function is increasing to the left of the vertex. Hence, the function is increasing on the interval \(( -\infty, \frac{3}{2})\).

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Key Concepts

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

Quadratic Functions
Quadratic functions are polynomial functions of degree two. They can be represented in the standard form: \[ f(x) = ax^2 + bx + c \] where:
  • a, b, and c are constants
  • x is the variable

The coefficient 'a' determines the direction of the parabola. If \( a > 0 \), the parabola opens upward. If \( a < 0 \), it opens downward.
The graph of a quadratic function is always a parabola. Understanding the shape and direction helps in analyzing the behavior of the function. The largest or smallest point on the graph is known as the vertex.
Vertex Formula
The vertex of a quadratic function is a crucial point and represents the maximum or minimum value of the function. To find the vertex of a quadratic function in standard form \( f(x) = ax^2 + bx + c \), we use the vertex formula. The vertex \((h, k)\) is given by:
  • \( h = -\frac{b}{2a} \)
  • \( k = f(h) \), so substitute \( h \) back into the function to find \( k \)

For the given function \( f(x) = -x^2 + 3x - 10 \), we identified \( a = -1 \), \( b = 3 \), and \( c = -10 \). Using the vertex formula: \[ h = -\frac{3}{2(-1)} = \frac{3}{2} \].
Substituting \( h \) back into the function to find \( k \): \[ k = f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) - 10 \] \[ k = -\frac{9}{4} + \frac{9}{2} - 10 = -\frac{31}{4} \] Thus, the vertex is at \( \left(\frac{3}{2}, -\frac{31}{4}\right) \).
Intervals of Increase and Decrease
A quadratic function will either increase on one interval and decrease on another, divided by the vertex. For the quadratic function \( f(x) = ax^2 + bx + c \):
  • If \( a > 0 \), the parabola opens upward, and the function decreases on \( (-\infty, h) \) and increases on \( (h, \infty) \)
  • If \( a < 0 \), the parabola opens downward, and the function increases on \( (-\infty, h) \) and decreases on \( (h, \infty) \)
For \( f(x) = -x^2 + 3x - 10 \), since \( a = -1 \) (i.e., \( a < 0 \)), the parabola opens downward. Thus, the function increases to the left of the vertex and decreases to the right of the vertex.
Hence, the function is increasing on the interval: \[ (-\infty, \frac{3}{2}) \].

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