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How do you find the vertex and the intercepts for #y=x^2-10x+20#?

Answer 1

Vertex is at #(5,-5) # y-intercept is at #(0,20)# x-intercepts are at #(2.764,0) and (7.236,0)#

#y=x^2-10x+20 = (x-5)^2 -25+20 =(x-5)^2-5 :.#Comparing with vertex form of equation; #y=a(x-h)^2+k# we get vertex at #(h,k)=(5,-5)#.To get y-intercept putting x=0 in the equation we get #y=20# and to find x-intercepts putting y=0 in the equation we get #(x-5)^2-5=0 or (x-5)^2 =5 or (x-5) = +-sqrt5 or x= 5+-sqrt5 or x= 2.764,7.236 :.#Vertex is at #(5,-5) # y-intercept is at #(0,20)# x-intercepts are at #(2.764,0) and (7.236,0)# graph{x^2-10x+20 [-40, 40, -20, 20]}[Ans]

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Answer 2

Hazel Anglin

To find the vertex of the quadratic function y = x^2 - 10x + 20, you can use the formula x = -b/(2a), where 'a' is the coefficient of the x^2 term and 'b' is the coefficient of the x term.

To find the x-intercepts, set y = 0 and solve for x. These are the points where the graph crosses the x-axis.

To find the y-intercept, set x = 0 and solve for y. This is the point where the graph crosses the y-axis.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.