Can you explain how to graph the quadratic function f(x) = 4x^2 + 2x - 3 and identify its vertex?

Answers:

• To graph the quadratic function ( f(x) = 4x^2 + 2x - 3 ) and identify its vertex, follow these steps:

  Step 1: Identify the coefficients

  In the quadratic function ( f(x) = ax^2 + bx + c ), identify the coefficients:

  • ( a = 4 )
  • ( b = 2 )
  • ( c = -3 )

  Step 2: Find the vertex

  The vertex of a quadratic function given by ( f(x) = ax^2 + bx + c ) can be found using the formula for the x-coordinate of the vertex: [ x = -\frac{b}{2a} ] Plugging in the values of ( a ) and ( b ): [ x = -\frac{2}{2 \cdot 4} = -\frac{2}{8} = -\frac{1}{4} ]

  To find the y-coordinate of the vertex, substitute ( x = -\frac{1}{4} ) back into the function: [ f\left(-\frac{1}{4}\right) = 4\left(-\frac{1}{4}\right)^2 + 2\left(-\frac{1}{4}\right) - 3 ] Calculating this step-by-step:

  1. ( \left(-\frac{1}{4}\right)^2 = \frac{1}{16} )
  2. ( 4 \cdot \frac{1}{16} = \frac{1}{4} )
  3. ( 2 \cdot \left(-\frac{1}{4}\right) = -\frac{1}{2} )
  4. Combine these: [ f\left(-\frac{1}{4}\right) = \frac{1}{4} - \frac{1}{2} - 3 = \frac{1}{4} - \frac{2}{4} - \frac{12}{4} = -\frac{13}{4} ]

  Thus, the vertex is at: [ \left(-\frac{1}{4}, -\frac{13}{4}\right) ]

  Step 3: Determine the direction of the parabola

  Since the coefficient ( a = 4 ) is positive, the parabola opens upwards.

  Step 4: Find the y-intercept

  The y-intercept occurs when ( x = 0 ): [ f(0) = 4(0)^2 + 2(0) - 3 = -3 ] So, the y-intercept is ( (0, -3) ).

  Step 5: Find additional points

  To plot the graph more accurately, it can be useful to find a couple of additional points. Choose values around the vertex, such as ( x = -1 ) and ( x = 0 ), and calculate their corresponding ( f(x) ) values:

  • For ( x = -1 ): [ f(-1) = 4(-1)^2 + 2(-1) - 3 = 4 - 2 - 3 = -1 \quad \Rightarrow \quad (-1, -1) ]
  • For ( x = 0 ): [ f(0) = -3 \quad \Rightarrow \quad (0, -3) ]
  • For ( x = 1 ): [ f(1) = 4(1)^2 + 2(1) - 3 = 4 + 2 - 3 = 3 \quad \Rightarrow \quad (1, 3) ]

  Step 6: Sketch the graph

  Plot the vertex ( \left(-\frac{1}{4}, -\frac{13}{4}\right) ), the y-intercept ( (0, -3) ), and additional points ( (-1, -1) ) and ( (1, 3) ) on a coordinate plane.

  Step 7: Draw the parabola

  Connect these points with a smooth, upward-opening curve to represent the graph of the quadratic function.

  Summary of Key Points

  • Vertex: ( \left(-\frac{1}{4}, -\frac{13}{4}\right) )
  • Y-intercept: ( (0, -3) )
  • The parabola opens upwards.

  This process will allow you to graph the function accurately and understand its key features!

• To graph the quadratic function ( f(x) = 4x^2 + 2x - 3 ) and identify its vertex, follow these steps:

  Step 1: Identify the coefficients.

  The general form of a quadratic function is given by:

  [ f(x) = ax^2 + bx + c ]

  For your function, the coefficients are:

  • ( a = 4 )
  • ( b = 2 )
  • ( c = -3 )

  Step 2: Find the vertex.

  The vertex of a quadratic function in the form ( f(x) = ax^2 + bx + c ) can be found using the formula for the x-coordinate of the vertex:

  [ x = -\frac{b}{2a} ]

  Plugging in your values:

  [ x = -\frac{2}{2 \cdot 4} = -\frac{2}{8} = -\frac{1}{4} ]

  Now, substitute ( x = -\frac{1}{4} ) back into the function to find the y-coordinate of the vertex:

  [ f\left(-\frac{1}{4}\right) = 4\left(-\frac{1}{4}\right)^2 + 2\left(-\frac{1}{4}\right) - 3 ] [ = 4\left(\frac{1}{16}\right) - \frac{1}{2} - 3 ] [ = \frac{4}{16} - \frac{1}{2} - 3 ] [ = \frac{1}{4} - \frac{2}{4} - \frac{12}{4} ] [ = \frac{1 - 2 - 12}{4} = \frac{-13}{4} ]

  Therefore, the vertex of the function is:

  [ \left(-\frac{1}{4}, -\frac{13}{4}\right) \quad \text{or approximately} \quad (-0.25, -3.25) ]

  Step 3: Determine the direction of the parabola.

  Since ( a = 4 ) (which is positive), the parabola opens upwards.

  Step 4: Find the y-intercept.

  To find the y-intercept, evaluate the function at ( x = 0 ):

  [ f(0) = 4(0)^2 + 2(0) - 3 = -3 ]

  So, the y-intercept is ( (0, -3) ).

  Step 5: Find the x-intercepts.

  To find the x-intercepts, set ( f(x) = 0 ):

  [ 4x^2 + 2x - 3 = 0 ]

  You can use the quadratic formula:

  [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

  Substituting in the values:

  [ x = \frac{-2 \pm \sqrt{(2)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} ] [ = \frac{-2 \pm \sqrt{4 + 48}}{8} ] [ = \frac{-2 \pm \sqrt{52}}{8} ] [ = \frac{-2 \pm 2\sqrt{13}}{8} ] [ = \frac{-1 \pm \sqrt{13}}{4} ]

  Thus, the two x-intercepts are ( \left(\frac{-1 + \sqrt{13}}{4}, 0\right) ) and ( \left(\frac{-1 - \sqrt{13}}{4}, 0\right) ).

  Step 6: Plot the points.

  1. Vertex: ( \left(-\frac{1}{4}, -\frac{13}{4}\right) )
  2. Y-intercept: ( (0, -3) )
  3. X-intercepts: Find numerical approximate values for ( \frac{-1 \pm \sqrt{13}}{4} ) which are approximately ( (0.9, 0) ) and ( (-2.9, 0) ).

  Step 7: Sketch the graph.

  1. Plot the vertex.
  2. Plot the y-intercept and the x-intercepts.
  3. Draw a smooth curve through these points to form the parabola, making sure the shape opens upward due to positive ( a ).

  Final graph

  The graph should show a parabola opening upwards, with the vertex at ( \left(-\frac{1}{4}, -\frac{13}{4}\right) ) and crossing the y-axis at ( (0, -3) ) and the x-axis at the two calculated intercepts.