Question 949386: Find the coordinates of the vertex of the graph of the given quadratic function.
h(x)=x^2 + x + 5
this is how I solved it:
x(x^-2x-15)
x(x-5)(x+3)=0
x=0,5,8,-3
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! h(x) = x^2 + x + 5
the vertex of a quadratic equation is given by the formula of x = -b/2a
in the quadratic equation of x^2 + x + 5, which is in standard form, and has to be in standard form in order for the formula to work, you get:
a = 1
b = 1
c = 5
a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant term.
in the formula of x = -b/2a, replace b with 1 and a with 1 to get:
x = -1/2
when x = -1/2, h(x) = x^2 + x + 5 becomes 1/4 - 1/2 + 5 which becomes h(x) = 4 and 3/4.
the coordinates of the vertex are (x,y) = (-1/4, 4 and 3/4).
the graph of the equation is shown below:
you can see from the graph that the vertex is at x = -1/2 and y = 4 and 3/4.
draw an imaginary vertical line through the vertex and the x-axis, and an imaginary horizontal line through the vertex and the y-axis, and it will be easier to see.
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