The equation is a quadratic equation in standard form, or #y = color(red)(a)x^2 + color(green)(b)x + color(blue)(c)#.
The vertex is the minimum or maximum point of a parabola . To find the #x# value of the vertex, we use the formula #x_v = -color(green)(b)/(2color(red)(a))#, where #x_v# is the x-value of the vertex.
We know that #color(red)(a = 1)# and #color(green)(b = -4)#, so we can plug them into the formula: #x_v = (-(-4))/(2(1)) = 4/2 = 2#
To find the #y#-value, we just plug in the #x# value back into the equation: #y = (2)^2 - 4(2) + 9#
Simplify: #y = 4 - 8 + 9#
#y = -4+9#
#y = 5#
Therefore, the vertex is at #(2, 5)#
#--------------------#
Now to find the intercepts.
The #x#-intercept is the value of #x# when #y# equals to zero.
To find it, just plug in #0# for #y# in the equation: #0 = x^2 - 4x + 9#
Since we cannot factor #x^2 - 4x + 9#, we will use the quadratic formula# x = (-color(green)(b) +- sqrt(color(green)(b)^2-4color(red)(a)color(blue)(c))) / (2color(red)(a)) #
We already know that #color(red)(a = 1)#, #color(green)(b = -4)#, and #color(blue)(c = 9)#, so let's plug them into the formula: #x = (-(color(green)(-4)) +- sqrt(color(green)((-4))^2 - 4(color(red)(1))(color(blue)(9))))/(2(color(red)(1))#
Now simplify: #x = (4 +- sqrt(16 - 36))/2#
#x = (4+-sqrt(-20))/2#
Since we cannot do a square root of a negative number (it becomes imaginary), that means there are no #x#-intercepts .
The #y#-intercept is the value of #y# when #x# equals to zero.
To find it, just plug in #0# for all the #x#'s in the equation: #y = (0)^2 - 4(0) + 9#
Simplify: #y = 0 - 0 + 9#
#y = 9#
Now we write it as a coordinate, so it becomes #(0, 9)#.
