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).