Chapter 11, Problem 23STP

To determine

To evaluate: the leading coefficient of the quadratic function when the function is written in standard form.

Answer to Problem 23STP

The leading coefficient is 6.

Explanation of Solution

Given:

The graph of a quadratic function passes through the points $\left(-1,3\right),\left(0,1\right)$ and $\left(1,11\right)$ .

Calculation: -

Let the general equation of quadratic equation be $a{x}^2+bx+c=p\left(x\right)$

Putting $\left(-1,3\right)$ ,

  $\begin{array}{l}p\left(x\right)=a{x}^2+bx+c\\ p\left(-1\right)=a{\left(-1\right)}^2+b\left(-1\right)+c\\ 3=a-b+c\end{array}$

Putting $\left(0,1\right)$ ,

  $\begin{array}{l}p\left(x\right)=a{x}^2+bx+c\\ p\left(0\right)=a{\left(0\right)}^2+b\left(0\right)+c\\ 1=c\end{array}$

Putting $\left(1,11\right)$ ,

  $\begin{array}{l}p\left(x\right)=a{x}^2+bx+c\\ p\left(1\right)=a{\left(1\right)}^2+b\left(1\right)+c\\ 11=a+b+c\end{array}$

Solving the above equation,

  $\begin{array}{l}a-b+c=3\\ a-b+1=3\\ a-b=2\\ \\ \\ a+b+c=11\\ a+b+1=11\\ a+b=10\\ \\ \\ a-b=2\\ a+b=10\\ a=6,b=4\end{array}$

So,

The quadratic function becomes $p\left(x\right)=6x^2+4x+1$ .

Hence,

The leading coefficient is 6.