How do you find the vertex and the intercepts for #f(x)=-2x^2+2x-3#?

Question

Kennedy Adams · Accepted Answer

#color(blue)(y_("intercept")=-3#
#color(blue)("Vertex"->(x,y)->(1/2,-5/2))#
#color(red)("As the graph is of shape "nn" and the vertex")#
#color(red)("is below the x-axis there are no x-intercepts")# Given:#" "y=-2x^2+2x-3#......................(1) Standard form:#" "y=ax^2+bx+c# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine y intercept")# #color(blue)(y_("intercept")=c=-3# (note: y intercept is at x=0 so #-2x^2=0" and "2x=0" leaving " y=-3# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine x-vertex")# The coefficient of #x^2# is negative so the graph is of general shape #nn# Thus the vertex is a maximum. Write #" "y=-2x^2+2x-3" as "y=-2(x^2color(red)( -x)) -3# Using the coefficient of #color(red)(-x -> -1)# Apply: #color(blue)(x_("vertex")=(-1/2)xx(color(red)(-1))=+1/2)# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine y-vertex")# Substitute #x=1/2# into equation (1) giving #y_("vertex")=-2(1/2)^2+2(1/2)-3# #color(blue)(y_("vertex")=-2/4+2/2-3 = -5/2# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Vertex"->(x,y)->(1/2,-5/2))# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(red)("As the graph is of shape "nn" and the vertex")# #color(red)("is below the x-axis there are no x-intercepts")#

Hazel Anglin · Answer

undefined To find the vertex and intercepts for $ f(x) = -2x^2 + 2x - 3 $, follow these steps:

1. **Vertex:**
   Use the formula $ x = -\frac{b}{2a} $ to find the x-coordinate of the vertex.
   Once you have the x-coordinate, substitute it into the function to find the y-coordinate.

2. **x-intercepts:**
   Set $ f(x) = 0 $ and solve for $ x $.

3. **y-intercept:**
   Substitute $ x = 0 $ into the function $ f(x) $.

Let's apply these steps:

1. **Vertex:**
   $ x = -\frac{2}{2 \times (-2)} = \frac{1}{2} $
   Now, substitute $ x = \frac{1}{2} $ into the function:
   $ f(\frac{1}{2}) = -2(\frac{1}{2})^2 + 2(\frac{1}{2}) - 3 = -\frac{7}{2} $
   So, the vertex is $ (\frac{1}{2}, -\frac{7}{2}) $.

2. **x-intercepts:**
   Set $ f(x) = 0 $:
   $ -2x^2 + 2x - 3 = 0 $
   Solve this quadratic equation for $ x $.

3. **y-intercept:**
   Substitute $ x = 0 $ into the function:
   $ f(0) = -2(0)^2 + 2(0) - 3 = -3 $

So, the vertex of the function is $ (\frac{1}{2}, -\frac{7}{2}) $, and you can solve for the x-intercepts by solving the quadratic equation, while the y-intercept is $ (0, -3) $.