Q57. - Algebra 1

Question

Graph the function.

$y=3x^2-12x+5$

Step-by-Step Solution

Answer

The graph of the function $y=3x^2-12x+5$ is

Step 1. Define the standard form of the quadratic function.

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where, $a\ne 0$ is called the standard form of the quadratic function.

Step 2. Define the vertex of the function y = a x 2 + b x + c .

For the function $y=a{x}^2+bx+c$,

(1) If $a>0$, then the function has the minimum value at $x=-\frac{b}{2a}$ and the vertex is located at the minimum point.

(2) If $a<0$, then the function has the maximum value at $x=-\frac{b}{2a}$ and the vertex is located at the maximum point.

 

Opens upward and has a minimum value at $x=-\frac{b}{2a}$, when.

Opens downward and has a maximum value at $x=-\frac{b}{2a}$,  when. 

Step 3. Define y -intercept of the function y = a x 2 + b x + c .

The $y$-intercept of the function $y=a{x}^2+bx+c$ is always at $c$

Step 4. Plot the graph of the function y = 3 x 2 &#8722; 12 x + 5 .

Compare the quadratic function $y=3x^2-12x+5$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=3,b=-12,c=5$

Substitute $a=3$ and $b=-12$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{-12}{2\left(3\right)}\right) \\ x=-\left(\frac{-12}{6}\right) \\ x=-\left(-2\right) \\ x=2 \end{gathered}$$

Since, $a>0$.

So, the graph opens upward and has a minimum value at $x=2$.

Since, the $y$-intercept is given by $c$.

So, the $y$-intercept is 5. 

The graph of the function $y=3x^2-12x+5$ is shown below.