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How do you find the vertex and the intercepts for #y=x^2+9x+8#?

Answer 1

Vertex is at #(-9/2,-49/4)#; y-intercept is at #(0,8)#; x-intercept at #(-1,0) & (-8,0)#

#y=x^2+9x+8 = (x+9/2)^2-81/4+8=(x+9/2)^2-49/4:.#Vertex is at #(-9/2,-49/4)# Putting #x=0# we get y-intercept as #y=8#; Putting #y=0# we get x-intercept as #x^2+9x+8=0 or (x+8)(x+1)=0:. x= -8 ; x=-1# graph{x^2+9x+8 [-40, 40, -20, 20]}[Ans]

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Answer 2

Jameson Allen

To find the vertex of the quadratic function (y = x^2 + 9x + 8), use the formula (x = -\frac{b}{2a}). Then, substitute the (x)-value into the equation to find the corresponding (y)-value. To find the (x)-intercepts, set (y = 0) and solve for (x). To find the (y)-intercept, plug in (x = 0) into the equation.

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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.