/*! 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 2 Sketch the graph of the function... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Sketch the graph of the function. Choose a scale that allows all relative extrema and points of inflection to be identified on the graph. \(y=2 x^{2}-4 x+1\)

Short Answer

Expert verified
The graph of the function \(y=2x^{2}-4x+1\) is a parabola opening upwards. The vertex of the graph is at point (1,-1) and the axis of symmetry is \(x=1\).

Step by step solution

01

Analyze the parabolic function

A parabolic function of form \(y=ax^{2}+bx+c\) has a vertex \((h,k)\) determined by the formula \(h= -b/2a\). With \(a=2\) and \(b=-4\), calculate the value of \(h=-(-4)/2*2 = 1\).
02

Identify Vertex point

Now, using \(h=1\), get the \(y\)-coordinate of which is the value of \(k\), by substituting \(h\) into the function: \(k=2*(1)^{2}-4*1+1 = -1\). The vertex of the function is \((1,-1)\).
03

Identify Axis of symmetry

The axis of symmetry of a parabola equation \(y = ax^{2} + bx + c\) is the vertical line \(x = h\). So, for this function, the axis of symmetry is \(x=1\).
04

Sketch the graph

Plot the points, make sure to mark the vertex and axis of symmetry. Start at the vertex (1,-1), and because the coefficient \(a\) of \(x^{2}\) is positive, the parabola opens upwards. Draw a smooth, u-shaped curve.

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