Consider the quadratic function f(x) = -x^2 + 1x + 12. Determine the following: (enter all numerical answers as integers, fractions, or decimals): The smallest -intercept is 2 = The largest x-intercept is z = The y-intercept is y = The vertex is ( The line of symmetry has the equation Preview

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Answer and Explanation:Given : The quadratic function [tex]f(x)=-x^2+x+12[/tex]To find : Determine the following ?Solution : The x -intercept are where f(x)=0,So, [tex]-x^2+x+12=0[/tex]Applying middle term split,[tex]-x^2+4x-3x+12=0[/tex][tex]-x(x-4)-3(x-4)=0[/tex][tex](x-4)(-x-3)=0[/tex][tex]x=4,-3[/tex]The x-intercepts are (4,0) and (-3,0).The smallest x-intercept is x=-3The largest x-intercept is x=4The y -intercept are where x=0,So, [tex]f(0)=-(0)^2+0+12[/tex][tex]f(0)=12[/tex]The y-intercept is y=12.The quadratic function is in the form, [tex]y=ax^2+bx+c[/tex] On comparing, a=-1 , b=1 and c=1 2The vertex of the graph is denote by (h,k) and the formula to find the vertex is For h, The x-coordinate of the vertex is given by, [tex]h=-\frac{b}{2a}[/tex] [tex]h=-\frac{1}{2(-1)}[/tex] [tex]h=\frac{1}{2}[/tex] For k, The y-coordinate of the vertex is given by, [tex]k=f(h)[/tex] [tex]k=-h^2+h+12[/tex] [tex]k=-(\frac{1}{2})^2+\frac{1}{2}+12[/tex] [tex]k=-\frac{1}{4}+\frac{1}{2}+12[/tex] [tex]k=\frac{-1+2+48}{4}[/tex] [tex]k=\frac{49}{4}[/tex] The vertex of the function is [tex](h,k)=(\frac{1}{2},\frac{49}{4})[/tex] The x-coordinate of the vertex i.e. [tex]x=-\frac{b}{2a}[/tex] is the axis of symmetry, So, [tex]x=-\frac{b}{2a}=\frac{1}{2}[/tex] (solved above) The axis of symmetry is [tex]x=\frac{1}{2}[/tex].