Question

did you find it hard to pick x - values which gave you a good picture of the last graph on page 33? this would have been easier if you had known where the graph crossed the x - axis — in other words, where f(x)=0. we can find the x - values at these points by solving the equation x² + 8x + 7 = 0.

f(x)=x² + 8x + 7
0=x² + 8x + 7
0=(x + 7)(x + 1)
x + 7 = 0 or x + 1 = 0
x=-7 or x=-1

this means the graph crosses the x - axis at (-7,0) and (-1,0). we would get a good picture of the graph by choosing numbers between -7 and -1 as the other x - values in our table.

find the points where the graph of each quadratic function crosses the x - axis. approximate the x - values if you need to. then find at least three other points on the curve and draw the graph.

f(x)=x² - 16x + 60

f(x)=x² - 6x + 4

Explanation:

Step1: Find \(x -\)intercepts for \(f(x)=x^{2}-16x + 60\)

Set \(f(x)=0\) and factor.

Step2: Determine other points for \(f(x)=x^{2}-16x + 60\)

Substitute \(x\) - values.

Step3: Find \(x -\)intercepts for \(f(x)=x^{2}-6x + 4\)

Use quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step4: Determine other points for \(f(x)=x^{2}-6x + 4\)

Substitute \(x\) - values.

Answer:

1. For \(f(x)=x^{2}-16x + 60\):

• The points where the graph crosses the \(x -\)axis are found by setting \(f(x)=0\), so \(x^{2}-16x + 60=0\). Factoring gives \((x - 6)(x - 10)=0\). Then \(x=6\) or \(x = 10\). The points are \((6,0)\) and \((10,0)\).
• Three other points:
• When \(x=7\), \(f(7)=7^{2}-16\times7 + 60=49-112 + 60=-3\). So the point is \((7,-3)\).
• When \(x=8\), \(f(8)=8^{2}-16\times8 + 60=64-128 + 60=-4\). So the point is \((8,-4)\).
• When \(x=9\), \(f(9)=9^{2}-16\times9 + 60=81-144 + 60=-3\). So the point is \((9,-3)\).

1. For \(f(x)=x^{2}-6x + 4\):

• Set \(f(x)=0\), and use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b=-6\), \(c = 4\).
• \(x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times4}}{2\times1}=\frac{6\pm\sqrt{36 - 16}}{2}=\frac{6\pm\sqrt{20}}{2}=\frac{6\pm2\sqrt{5}}{2}=3\pm\sqrt{5}\).
• \(x_1=3+\sqrt{5}\approx3 + 2.24=5.24\) and \(x_2=3-\sqrt{5}\approx3 - 2.24 = 0.76\). The points are \((3+\sqrt{5},0)\approx(5.24,0)\) and \((3-\sqrt{5},0)\approx(0.76,0)\).
• Three other points:
• When \(x = 3\), \(f(3)=3^{2}-6\times3 + 4=9-18 + 4=-5\). So the point is \((3,-5)\).
• When \(x=4\), \(f(4)=4^{2}-6\times4 + 4=16-24 + 4=-4\). So the point is \((4,-4)\).
• When \(x=2\), \(f(2)=2^{2}-6\times2 + 4=4-12 + 4=-4\). So the point is \((2,-4)\).