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How do you find the vertex and the intercepts for #y=x^2+7x#?

Answer 1

Eva Adamson

#"see explanation"#

#"for intercepts"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercepts"#

#x=0toy=0larrcolor(red)"y-intercept"#

#y=0tox^2+7x=0#

#rArrx(x+7)=0#

#rArrx=0" or "x=-7larrcolor(red)"x-intercepts"#

#"the vertex lies on the axis of symmetry which is positioned"# #"at the midpoint of the x-intercepts"#

#x_(color(red)"vertex")=(0-7)/2=-7/2#

#"substitute this value into the equation for y"#

#y_(color(red)"vertex")=(-7/2)^2+7(-7/2)=-49/4#

#rArrcolor(magenta)"vertex "=(-7/2,-49/4)# graph{x^2+7x [-10, 10, -5, 5]}

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Answer 2

Nathan Axel

To find the vertex of the quadratic function ( y = x^2 + 7x ), use the formula ( x = \frac{{-b}}{{2a}} ), where ( a ) is the coefficient of the ( x^2 ) term and ( b ) is the coefficient of the ( x ) term.

For the intercepts:

To find the x-intercepts, set ( y = 0 ) and solve for ( x ).
To find the y-intercept, set ( x = 0 ) and solve for ( y ).

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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.