Question 88766
Find the vertex of the parabola defined by:
{{{f(x) = 2x^2+4x-15}}} Compare this with the general form:
{{{f(x) = ax^2+bx+c}}}...and you can see that: a = 2, b = 4, and c = -15
Now substitute a and b into the formula for the x-coordinate of the vertex:{{{-b/2a}}}
{{{x = -4/2(2)}}}
{{{x = -4/4}}}
{{{x = -1}}} This is the x-coordinate of the vertex.  Substitute this value of x into the given quadratic equation and solve for f(x), this is the same as the y-coordinate.
{{{f(-1) = 2(-1)^2+4(-1)-15}}} Evaluate.
{{{f(-1) = 2-4-15}}}
{{{f(-1) = -17}}}
The coordinates of the vertex are:
(-1, -17)...and the equation of the line of symmetry is:{{{x = -1}}}
The graph looks like this:
{{{graph(400,400,-5,5,-20,5,2x^2+4x-15)}}}
To plot two points to the left and to the right of the vertex (line of symmetry), choose four values of x-coordinate, such as:
x = 0
x = -2
x = 1
x = -3
and for each one of these, find the corresponding y-coordinate (f(x)) and these will give you the coordinates of the four points.
I'll do one and you can finish the rest.
Substitute x = -3
{{{f(-3) = 2(-3)^2+4(-3)-15}}} Evaluate.
{{{f(x) = 2(9)+(-12)-15}}}
{{{f(x) = 18-12-15}}}
{{{f(x) = -9}}}
The coordinates of this point are: (-3, -9)