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Grade 9 · unit review

Review: Algebra1: Quadratic Formula and Graph Features

Quadratic equations and graphs express the same underlying structure. A discriminant predicts real-root count, the quadratic formula calculates roots, and vertex form displays shifts, direction and stretching directly.

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Unit review questions
1. Solve 2x²−30x+112=0 using the quadratic formula.
Need a hint?
  • For Ax²+Bx+C=0 with A nonzero, roots are [−B±√(B²−4AC)]/(2A).
  • The entire numerator is divided by 2A.
2. How many real roots does x²+14x+40=0 have?
Need a hint?
  • The discriminant D=B²−4AC is the expression under the formula’s square root.
  • A negative discriminant means no real roots, not no complex roots.
3. Find the vertex of y=(x−7)²−8.
Need a hint?
  • In y=(x−h)²+k, the square is nonnegative and becomes zero at x=h.
  • The sign inside x−h is opposite the horizontal coordinate’s written sign.
4. Compare y=−2(x−7)²+8 with y=x².
Need a hint?
  • The coefficient outside a squared expression controls output sign and vertical scale.
  • A negative leading coefficient makes the graph open downward.

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Use the skills in a new task

Create a four-part mathematical solution sheet for these supplied cases. 1. Solve 2x²−38x+180=0 using the quadratic formula. Candidate answers: (A) x=18 or x=20 (B) x=19 only (C) x=9 or x=10 (D) x=−9 or x=−10. 2. How many real roots does x²+18x+83=0 have? Candidate answers: (A) Every real number is a root (B) No real roots (C) One repeated real root (D) Two distinct real roots. 3. Find the vertex of y=(x−9)²−10. Candidate answers: (A) (0,71) (B) (9,10) (C) (9,−10) (D) (−9,−10). 4. Compare y=−2(x−9)²+10 with y=x². Candidate answers: (A) Only shift left and down (B) Open upward and compress vertically by 2 (C) Reflect across x-axis, stretch vertically by 2, shift right and up (D) Reflect across y-axis, then shift left.

Work on paper or in your own document. Explain the choices you made.

  • State each requested quantity or relationship and its input restrictions.
  • Show intermediate calculations or the applicable theorem with matching correspondences.
  • Give all four results, including exact values or requested rounding and units.
  • Check a result by substitution, reversal, dimensions, a representation or a counterexample as appropriate.
Compare with a model response

1. x=9 or x=10. The discriminant is [-38]²−4(2)(180)=4. Thus x=[38±2]/4, giving 9 and 10. Keep the full numerator over 2A. 2. No real roots. D=(18)²−4(83)=-8. A positive discriminant gives two distinct real roots; zero gives one repeated real root; a negative discriminant gives no real roots. Here the answer is: no real roots. 3. (9,−10). In y=(x−h)²+k, the square is least at x=h. Here h=9 and k=−10, so the vertex is (9,−10). 4. Reflect across x-axis, stretch vertically by 2, shift right and up. The leading −2 reverses the output sign and doubles its magnitude. The input x−9 shifts right 9; +10 shifts up 10. The vertex is (9,10).

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