Quadratic Graphs

Step 1: The coefficient of $x^2$ is positive, so the graph is u-shaped.

Step 2: When $y=0$,

$begin{aligned} 2x^2 + 3x – 2 &= 0 (2x-1)(x+2) &= 0 end{aligned}$

$x = dfrac{1}{2}$  or  $x = -2$

Hence, the curve crosses the $x$-axis at $textcolor{red}{left( dfrac{1}{2}, 0 right)}$ and $textcolor{red}{(-2,0)}$

When $x = 0$,

$y = 2(0)^2 + 3(0) – 2 = -2$

Hence, the curve crosses the $y$-axis at $textcolor{red}{(0,-2)}$

Note: You can often read off the value the graph crosses the $y$-axis at (called the $y$-intercept) from the equation of a quadratic. The $y$-intercept of $y=ax^{2}+bx+c$ is always $c$.

Step 3: The minimum value will be halfway between $left(dfrac{1}{2}, 0 right)$ and $(-2,0)$:

$left( dfrac{1}{2} – 2 right) div 2 = – dfrac{3}{4}$

So, the minimum value is at $x = – dfrac{3}{4}$

Therefore, the minimum is

$y = 2 left(- dfrac{3}{4} right)^2 + 3 left(- dfrac{3}{4} right) – 2 = – dfrac{25}{8}$

So, the graph has a minimum at the point $textcolor{red}{left(- dfrac{3}{4}, – dfrac{25}{8} right)}$

Step 4: Sketch the graph (sketching doesn’t need to be to scale) and label all the important points.

Firstly, complete the square:

$y = 3x^2 – 12x + 7 = 3(x-2)^2 – 5$

When $y=0$,

$begin{aligned} 3(x-2)^2 – 5 &=0 (x-2)^2 &= dfrac{5}{3} x – 2 &= pm sqrt{dfrac{5}{3}} x &= 2 pm dfrac{sqrt{15}}{3} end{aligned}$

So, the curve crosses the $x$-axis at $textcolor{red}{left(2-dfrac{sqrt{15}}{3},0 right)}$ and $textcolor{red}{left(2+dfrac{sqrt{15}}{3},0 right)}$

When $x=0$,

$y = 3(0)^2 – 12(0) + 7 = 7$

So, the curve crosses the $y$-axis at $textcolor{red}{(0,7)}$

$(x-2)^2 geq 0$ (since a square number is never less than $0$), therefore the minimum value occurs at $x-2=0 Rightarrow x = 2$, and therefore the minimum is

$y = 3(2-2)^2 – 5$

So, the minimum is at the point $textcolor{red}{(2, -5)}$

Then, you have enough information to sketch the graph.

Note: You can read off the vertex of a graph from completed square form. The vertex of $y=a(x+b)^{2}+c$ is always at $(-b,c)$.