Respuesta :
11. The given parabola has equation: [tex]f(x)=-3x^2-18x-35[/tex]
We complete the square to have the function in the vertex form: [tex]f(x)=-3(x^2+6x)-35[/tex]
[tex]f(x)=-3(x^2+6x+9)+3*9-35[/tex]
[tex]f(x)=-3(x^2+6x+9)+27-35[/tex]
[tex]f(x)=-3(x+3)^2-8[/tex]
The vertex is (-3,-8), the graph opens down because a=-3<0
Ans: B. Vertex: (−3, −8), opens downward
12. We know the graph of [tex]y=x^2[/tex] is a parabola that opens up and has its minimum point at the origin.
The range of this function is [tex]y\ge 0[/tex]
A straight line that has its y-intercept below the x-axis (y=3x-6) will not intersect [tex]y=x^2[/tex] and hence will form a system with no solution.
Ans: D. y equals x squared, y equals 3x minus 6
14. We want to find the zeros of [tex]x^2+21x=5x-63[/tex]
[tex]x^2+21x-5x+63=0[/tex]
[tex]x^2+16x+63=0[/tex]
The factored form is [tex](x+7)(x+9)=0[/tex]
[tex]x=-7\:or\: x=-9[/tex]
Ans: C. x = −7, −9
15. The given quadratic expression is [tex]6x^2-17x+10[/tex]
We split the middle term to obtain:
[tex]6x^2-12x-5x+10=0[/tex]
We factor by grouping to get:
[tex]6x(x-2)-5(x-2)=0[/tex]
[tex](x-2)(6x-5)=0[/tex]
Ans: D. 6x − 5
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