The solutions, or roots, of ax^2+bx+c=0 are the x-intercepts of the graph of f(x)=ax^2+bx+c. Let's write our equation in standard form. This means gathering all of the terms on the left-hand side of the equation.
-3x^2 + 2x = 15
⇕
-3x^2 + 2x - 15 = 0
Now we can identify the function related to the equation.
To draw the graph of the related quadratic function written in standard form, we must start by identifying the values of a, b, and c.
f(x)= - 3x^2+2x-15=0
⇕
f(x)= - 3x^2+ 2x+( - 15)
We can see that a= - 3, b= 2, and c= - 15. Now, we will follow three steps to graph the function.
The axis of symmetry of the parabola is the vertical line with equation x≈0.33.
Making the Table of Values
Next, we will make a table of values using x- values around the axis of symmetry x≈0.33.
x
- 3x^2+2x-15
f(x)
- 0.33
- 3( - 0.33)^2+2( - 0.33)-15
- 16
0
- 3( 0)^2+2( 0)-15
- 15
0.33
- 3( 0.33)^2+2( 0.33)-15
- 14.66
0.66
- 3( 0.66)^2+2( 0.66)-15
- 15
1
- 3( 1)^2+2( 1)-15
- 16
Plotting and Connecting the Points
We can finally draw the graph of the function. Since a=- 3, which is negative, the parabola will open downwards. Let's connect the points with a smooth curve.
Finding the x-intercepts
Let's identify the x-intercepts of the graph of the related function.
We can see that the parabola does not intersect the x-axis. Therefore, the equation - 3x^2+2x-15=0 does not have any real solutions.
