Question 91110
To find the y-intercept, plug in x=0


Lets find the value of y when  {{{x=0}}}


{{{y=x^2-3x-4}}} Start with the given polynomial



{{{y=(0)^2-3(0)-4}}} Plug in {{{x=0}}}



{{{y=(0)-3(0)-4}}} Raise 0 to the second power to get 0



{{{y=(0)-0-4}}} Multiply 3 by 0 to get 0



{{{y=-4}}} Remove any zero terms



So when {{{x=0}}}, {{{y=-4}}}


So the y-intercept is (0,-4)



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To determine the x-value of the vertex, use this formula

{{{x=-b/(2a)}}}


From the equation {{{y=x^2-3x-4}}} we can see that a=1 and b=-3


{{{x=(--3)/(2*1)}}} Plug in b=-3 and a=1



{{{x=3/(2*1)}}} Negate -3 to get 3



{{{x=(3)/2}}} Multiply 2 and 1 to get 2


{{{x=3/2}}} Reduce




So the x-coordinate of the vertex is {{{x=3/2}}} (which is {{{x=1.5}}} in decimal form). Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(1.5)}}}


{{{f(x)=x^2-3x-4}}} Start with the given polynomial



{{{f(1.5)=(1.5)^2-3(1.5)-4}}} Plug in {{{x=1.5}}}



{{{f(1.5)=2.25-3(1.5)-4}}} Raise 1.5 to the second power to get 2.25



{{{f(1.5)=2.25-4.5-4}}} Multiply -3 and 1.5 to get -4.5



{{{f(1.5)=-6.25}}} Now combine like terms



So the vertex is (1.5,-6.25)



Here is a graph to see the vertex and y-intercept 


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-3x-4) }}}