Question

the exponential grows at half the rate of the quadratic. the exponential grows at the same rate as the quadratic. the exponential grows at twice the rate of the quadratic. the exponential grows at four times the rate of the quadratic.

Explanation:

Step1: Identify the functions

The quadratic function (parabola) passes through \((0,0)\) and \((1,1)\), so its equation is \(y = x^2\) (since for \(y = ax^2+bx + c\), \(c = 0\), \(b = 0\), and when \(x = 1\), \(y = 1\) gives \(a = 1\)). The exponential function passes through \((0,1)\) and \((1,2)\), so its equation is \(y = 2^x\) (since for \(y = a\cdot b^x\), \(a = 1\) (when \(x = 0\), \(y = 1\)) and when \(x = 1\), \(y = 2\) gives \(b = 2\)).

Step2: Calculate the rate of change

For the quadratic function \(y=x^2\), the rate of change (average rate of change) over \([0,1]\) is \(\frac{f(1)-f(0)}{1 - 0}=\frac{1 - 0}{1}=1\).

For the exponential function \(y = 2^x\), the rate of change over \([0,1]\) is \(\frac{f(1)-f(0)}{1 - 0}=\frac{2 - 1}{1}=1\)? Wait, no, wait. Wait, the exponential at \(x = 0\) is \(1\), at \(x = 1\) is \(2\). The quadratic at \(x = 0\) is \(0\), at \(x = 1\) is \(1\). Wait, the rate of the exponential: \(\frac{2 - 1}{1-0}=1\)? No, that can't be. Wait, maybe I misread the graph. Wait, the exponential passes through \((0,1)\) and \((1,2)\), so change is \(2 - 1=1\) over interval length \(1\). The quadratic passes through \((0,0)\) and \((1,1)\), change is \(1 - 0 = 1\) over interval length \(1\). Wait, but that would mean same rate? But wait, maybe the exponential at \(x = 1\) is \(2\) and quadratic at \(x = 1\) is \(1\). Wait, the rate of growth: the exponential's rate (average rate) is \(\frac{2 - 1}{1}=1\), quadratic's rate is \(\frac{1 - 0}{1}=1\). Wait, but that would mean same rate? But let's check again. Wait, the exponential function: when \(x = 0\), \(y = 1\); \(x = 1\), \(y = 2\). Quadratic: \(x = 0\), \(y = 0\); \(x = 1\), \(y = 1\). So the change in exponential is \(2 - 1 = 1\), change in quadratic is \(1 - 0 = 1\). So the rate (average rate of change) is the same. Wait, but maybe the question is about the rate of growth (instantaneous? No, over the interval \([0,1]\)). So the average rate of change for quadratic is \(1\), for exponential is \(1\)? Wait, no, wait, the exponential at \(x = 1\) is \(2\), quadratic at \(x = 1\) is \(1\). So the exponential grows from \(1\) to \(2\) (change of \(1\)) and quadratic grows from \(0\) to \(1\) (change of \(1\)) over the same interval. So their rates (average rates) are the same.

Wait, but maybe I made a mistake. Let's re - evaluate. The exponential function: at \(x = 0\), \(y = 1\); at \(x = 1\), \(y = 2\). So the amount of growth is \(2 - 1 = 1\). The quadratic function: at \(x = 0\), \(y = 0\); at \(x = 1\), \(y = 1\). Amount of growth is \(1 - 0 = 1\). So over the interval \([0,1]\), both have a rate of growth (average rate of change) of \(1\) per unit \(x\). So the exponential grows at the same rate as the quadratic.

Answer:

The exponential grows at the same rate as the quadratic.