Problem 65 - Algebra 1: Concepts and Skills

Question

Sketch the graph of the function. $$y=4 x^{2}-x+6$$

Step-by-Step Solution

Answer

The graph of the function \(y=4x^2-x+6\) is a parabola that opens upwards, with vertex at \((1/8, 6-1/8)\) and the axis of symmetry as the line \(x=1/8\).

Step 1: Determine the direction of opening of the parabola

Since the coefficient of \(x^2\) is positive (4), the parabola opens upwards.

Step 2: Find the vertex and the axis of symmetry

The vertex \((h, k)\) of a parabola given by the equation \(y=a(x-h)^2 + k\) (where \(a\) is the coefficient of \(x^2\)) is given by the formula \(h= -b/2a\), where \(b\) is the coefficient of \(x\) and \(a\) is the coefficient of \(x^2\). Here \(a=4\) and \(b=-1\). So, \(h= -(-1)/(2*4)=1/8\). Substitute \(h=1/8\) in the equation to get \(k\). The axis of symmetry is the line \(x=h\).

Step 3: Graph the function

Plot the vertex, draw the axis of symmetry, and then draw a parabola that opens upwards and is symmetric around the axis of symmetry.