,"correctAnswer":"-5"}],"sentence":"For $y=4(x-3)^2-5$, the turning point is ({{blank:x}}, {{blank:y}})."},{"id":"card-9","type":"content","order":9,"title":"Parameters and edge cases","blocks":[{"id":"block-1","content":"When a parameter appears in the coefficient of $x^2$, first check that the equation is still quadratic. If that coefficient becomes 0, the discriminant rules for quadratics no longer apply."},{"id":"block-2","content":"For $(k+1)x^2+2x+k=0$ to have two distinct real roots, first require $k\\neq-1$ so the equation stays quadratic. Then use $\\Delta>0$:\n$$2^2-4(k+1)k>0$$\nThis simplifies to the quadratic inequality $k^2+k-1<0$."},{"id":"block-3","content":"Solve the boundary equation $k^2+k-1=0$ to find the critical values:\n$$k=\\frac{-1\\pm\\sqrt{5}}{2}$$\nThe quadratic in $k$ is negative between these roots, meaning $k$ must lie within that range."},{"id":"block-4","content":"The final solution is:\n$$\\frac{-1-\\sqrt{5}}{2}\\frac{-1+\\sqrt{5}}{2}$","isCorrect":false},{"id":"c","text":"$$\\frac{-1-\\sqrt{5}}{2}0$ gives the interval between the two roots of $k^2+k-1$, but $k=-1$ must be excluded since the equation would no longer be quadratic. A is the most common wrong answer because it forgets to remove that excluded value.","correctOptionId":"c"},{"id":"card-11","type":"flashcard","cards":[{"id":"fc-1","back":"$$\\Delta>0$","front":"Two distinct real roots"},{"id":"fc-2","back":"$$\\Delta=0$","front":"Repeated root or tangency"},{"id":"fc-3","back":"$$\\Delta<0$","front":"No real roots"},{"id":"fc-4","back":"$$(h,k)$","front":"Turning point of $y=a(x-h)^2+k$"}],"order":11},{"id":"card-12","type":"content","order":12,"title":"Tangency and repeated roots","blocks":[{"id":"block-1","content":"When a line is tangent to a curve, the two graphs meet at one point. After rearranging, the resulting equation in $x$ has a repeated root, so its discriminant is zero."},{"id":"block-2","content":"For the curve $y=\\frac{3}{x}+k$ and the line $y=-x+4$, set them equal and multiply by $x$ to clear the denominator:\n$$\\frac{3}{x}+k=-x+4 \\Rightarrow 3+kx=-x^2+4x$$"},{"id":"block-3","content":"Rearranging the equation into a standard quadratic form gives:\n$$x^2+(k-4)x+3=0$$\nNow we can apply the condition for tangency by setting the discriminant $\\Delta = 0$."},{"id":"block-4","content":"$$$(k-4)^2-4(1)(3)=0$$\nSolving for $k$, we get $(k-4)^2=12$, which simplifies to $k=4\\pm2\\sqrt{3}$. These are the two values of $k$ for which the line just touches the curve."}]},{"id":"card-13","type":"fill_blank","order":13,"blanks":[{"id":"k","isMath":true,"correctAnswer":"\\frac{5}{4}"}],"sentence":"If the line $y=x+1$ is tangent to the curve $y=x^2+k$, then $k =$ {{blank:k}}."},{"id":"card-14","type":"content","order":14,"title":"Quadratic models","blocks":[{"id":"block-1","content":"In modelling questions, completed-square form makes the maximum or minimum easy to spot. For $P=50-2(x-12)^2$, the square term is always non-negative, so the largest possible value occurs when $(x-12)^2=0$."},{"id":"block-2","content":"That gives a maximum profit of £50,000 when $x=12$, because $P$ is measured in thousands of pounds. To find when profit is more than £32,000, solve\n\n$$50-2(x-12)^2>32$$"},{"id":"block-3","content":"Rearranging gives\n\n$$(x-12)^2<9$$\n\nso\n\n$$-3

Quadratics Lesson

Quadratics Lesson