Find the value of $k$ for which the curve $y = 2x^2 - 3x + k$ (i) passes through the point $(4, -7)$, (ii) meets the $x$-axis at one point only.

Additional Mathematics 4037/11 — May/June 2016

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

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Topics Calculus · Trigonometry · Quadratic functions · Logarithmic and exponential functions · [Legacy] Indices and surds · [Legacy] Matrices · +6 more

Q1Quadratic functionsFree sample

Find the value of $k$ for which the curve $y = 2x^2 - 3x + k$

(i)

passes through the point $(4, -7)$ ,

DifficultyEasy

Worked solution

Approach

The curve passes through $(4, -7)$ , so substitute $x = 4$ and $y = -7$ into $y = 2x^2 - 3x + k$ and solve for $k$ .

Working

-7 = 2(4)^2 - 3(4) + k

-7 = 32 - 12 + k = 20 + k

k = -7 - 20 = -27

Answer

k = -27

Final answer

-27

Detailed explanation

Walkthrough

Since the point $(4, -7)$ lies on the curve, its coordinates must satisfy the curve's equation. Substituting $x = 4$ gives $2(4)^2 = 32$ and $3(4) = 12$ , so the equation becomes $-7 = 20 + k$ . Rearranging gives $k = -27$ .

Key Takeaways

A point lies on a curve exactly when its coordinates satisfy the curve's equation; substituting the point turns the unknown constant into a solvable linear equation.

Common Mistakes

Arithmetic slips in evaluating $2(4)^2$ (it is $32$ , not $16$ or $64$ ) or sign errors when rearranging $-7 = 20 + k$ .

Things to Be Careful About

Show the substituted equation before the final value — the single B1 is awarded for the correct value reached from valid working.

Techniques used

substitute the coordinates of the point into the curve equationsolve the resulting linear equation for k

(ii)

meets the $x$ -axis at one point only.

DifficultyMedium-Easy

Worked solution

Approach

The curve meets the $x$ -axis where $2x^2 - 3x + k = 0$ . Meeting the axis at exactly one point means this quadratic has a repeated root, so its discriminant is zero.

Working

For $2x^2 - 3x + k = 0$ , we have $a = 2$ , $b = -3$ , $c = k$ .

b^2 - 4ac = (-3)^2 - 4(2)(k) = 9 - 8k

Setting the discriminant equal to zero:

9 - 8k = 0

k = \frac{9}{8}

Answer

k = \frac{9}{8}

Final answer

k = 9/8

Detailed explanation

Walkthrough

A curve meets the $x$ -axis where $y = 0$ , giving the quadratic equation $2x^2 - 3x + k = 0$ . Meeting the axis at one point only means the quadratic has exactly one (repeated) root, which happens precisely when the discriminant $b^2 - 4ac$ equals zero. Here $b^2 = (-3)^2 = 9$ and $4ac = 8k$ , so $9 - 8k = 0$ , giving $k = \frac{9}{8}$ . (Equivalently, one could complete the square to get $y = 2\left(x - \frac{3}{4}\right)^2 + k - \frac{9}{8}$ and set the constant term to zero, or set the derivative $4x - 3$ equal to zero at the touching point — all routes give the same answer.)

Key Takeaways

The condition 'touches the $x$ -axis at one point' translates to a zero discriminant, $b^2 - 4ac = 0$ ; this is the standard link between the geometry of a parabola and the algebra of its roots.

Common Mistakes

Sign slips in $b^2 - 4ac$ (forgetting $(-3)^2 = +9$ ); using $b^2 - 4ac > 0$ (two intersections) or $< 0$ (no intersections) instead of $= 0$ ; forgetting to set the discriminant to zero and leaving $9 - 8k$ unevaluated.

Things to Be Careful About

The M1 requires a complete method reaching a value of $k$ — show the equation $9 - 8k = 0$ explicitly before the final answer, and give the exact fraction $\frac{9}{8}$ .

Techniques used

set the discriminant of the quadratic to zerosolve for the value of k giving a repeated root

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