Question

Here is a sketch of a curve.

The equation of the curve is y = x² + ax + b
where a and b are integers.
The points (0, -7) and (7, 0) lie on the curve.
Find the coordinates of the turning point of the curve.
Finish your answer by writing, Turning point at (..., ...)
YA
O
+

Answer 1

`{\Large \begin{array}{llll} y=x^2+ax+b \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=0\\ y=-7 \end{cases}\implies -7=0^2+a(0)+b\implies \boxed{-7=b} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} x=7\\ y=0 \end{cases}\implies 0=7^2+a7+\stackrel{b}{(-7)}\implies 0=49+7a-7 \\\\\\ 0=42+7a\implies -42=7a\implies \cfrac{-42}{7}=a\implies \boxed{-6=a} \\\\\\ ~\hfill {\Large \begin{array}{llll} y=x^2-6x-7 \end{array}}~\hfill`

now, let's get the "vertex" using the coefficients.

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{-6}x\stackrel{\stackrel{c}{\downarrow }}{-7} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ -6}{2(1)}~~~~ ,~~~~ -7-\cfrac{ (-6)^2}{4(1)}\right) \implies \left( - \cfrac{ -6 }{ 2 }~~,~~-7 - \cfrac{ 36 }{ 4 } \right) \\\\\\ \left( 3 ~~~~ ,~~~~ -7 -9 \right)\implies {\Large \begin{array}{llll} (3~~,~-16) \end{array}}`