How do you sketch this...I'm a bit confused since it's quadratic (I need help on the full method of the intercepts and turning point) h(x) = 3 {x}^(2) + 12x - 12 ​

Question

How do you sketch this...I'm a bit confused since it's quadratic (I need help on the full method of the intercepts and turning point)


`h(x) = 3 {x}^(2) + 12x - 12`
​

Answer 1

to sketch it we can rely on the vertex and the zeros, in this case the zeros are -2±2√2, which loosely give us the about 0.828 and -4.828, now let's find the vertex

`\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{3}x^2\stackrel{\stackrel{b}{\downarrow }}{+12}x\stackrel{\stackrel{c}{\downarrow }}{-12} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 12}{2(3)}~~~~ ,~~~~ -12-\cfrac{ (12)^2}{4(3)}\right) \implies\left( - \cfrac{ 12 }{ 6 }~~,~~-12 - \cfrac{ 144 }{ 12 } \right) \\\\\\ \left( - \cfrac{ 12 }{ 6 }~~,~~-12 - 12 \right)\implies (-2~~,~-24)`

Check the picture below.

since quadratics are just parabolas and are symmetric, so from the vertex to one of the zeros you can draw a hump or curve and the same curve across the axis of symmetry.