Skip to main content
뒤로

Precalculus Midterm 2 Practice Flashcards

컨트롤 버튼이 '내비게이션' 모드로 변경되었습니다.
1/14
  • Write the quadratic function q(x) = x² − 10x + 3 in standard form.
    Standard form is \(a(x-h)^2+k\). Complete the square to get \((x-5)^2-22\).
  • Solve the inequality 3x + 5 / (x + 9) ≥ 0 in interval form.
    Find zeros and undefined points: numerator zero at x = -5/3, denominator zero at x = -9. Test intervals to get solution (-9, -5/3].
  • Solve the inequality x² + 3x − 4 ≤ 0 in interval form.
    Factor as \((x+4)(x-1)\). Inequality holds between roots, so solution is [-4, 1].
  • Find the vertex of q(x) = x² − 6x − 7.
    Vertex form: complete the square or use \(h=-\frac{b}{2a}\). Vertex is (3, -16).
  • Find the x-intercepts of q(x) = x² − 6x − 7.
    Solve \(x^2-6x-7=0\). Roots are x = 7 and x = -1. Intercepts are points (7,0) and (-1,0).
  • Calculate the horizontal asymptote(s) of R(x) = (x + 213)(x + 51) / (x + 20)(x − 12).
    Degrees of numerator and denominator are equal. Horizontal asymptote is ratio of leading coefficients: y = 1.
  • Calculate the vertical asymptote(s) of R(x) = (x + 213)(x + 51) / (x + 20)(x − 12).
    Vertical asymptotes occur where denominator is zero: x = -20 and x = 12.
  • Explain what R(−10,000) is approximately equal to without calculating.
    For large |x|, R(x) ≈ ratio of leading terms, so R(−10,000) ≈ 1.
  • Why is the horizontal asymptote of h(x) = (666x − 5.6) / (x² + 1) equal to y = 0?
    Degree of denominator (2) is greater than numerator (1), so horizontal asymptote is y = 0.
  • Find the standard form of a parabola with vertex (1,1) passing through (2,2).
    Use vertex form \(y=a(x-1)^2+1\). Substitute (2,2) to find a=1. Equation: \(y=(x-1)^2+1\).
  • Calculate the base b that gives maximum area for A(b) = 26b − b².
    Maximum at vertex: \(b=\frac{-26}{-2} = 13\).
  • Calculate the maximum area for A(b) = 26b − b² when b = 13.
    Substitute b=13: \(A(13)=26(13)-13^2=169\).
  • Explain why the calculated area at b=13 is a maximum.
    The quadratic has a negative leading coefficient (-1), so the parabola opens downward, making the vertex a maximum.
  • Describe a quadratic polynomial with no x-intercepts.
    A parabola that does not cross the x-axis has a negative discriminant and lies entirely above or below the x-axis.