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 $f(x)=0$, so $x^{2}-16x + 60=0$. Factor the quadratic: $(x - 6)(x - 10)=0$. Then $x-6 = 0$ or $x - 10=0$, giving $x=6$ or $x = 10$. The x - intercepts are $(6,0)$ and $(10,0)$.

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

Let $x=5$, then $f(5)=5^{2}-16\times5 + 60=25-80 + 60=5$.
Let $x=7$, then $f(7)=7^{2}-16\times7 + 60=49-112 + 60=-3$.
Let $x=11$, then $f(11)=11^{2}-16\times11 + 60=121-176+60=5$.

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

Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c=0$. Here $a = 1$, $b=-6$, $c = 4$. Then $x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times4}}{2\times1}=\frac{6\pm\sqrt{36 - 16}}{2}=\frac{6\pm\sqrt{20}}{2}=\frac{6\pm2\sqrt{5}}{2}=3\pm\sqrt{5}$. The x - intercepts are $(3+\sqrt{5},0)\approx(5.24,0)$ and $(3 - \sqrt{5},0)\approx(0.76,0)$.

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

Let $x=3$, then $f(3)=3^{2}-6\times3 + 4=9-18 + 4=-5$.
Let $x=1$, then $f(1)=1^{2}-6\times1 + 4=1-6 + 4=-1$.
Let $x=6$, then $f(6)=6^{2}-6\times6 + 4=4$.

Answer:

For $f(x)=x^{2}-16x + 60$, x - intercepts are $(6,0)$ and $(10,0)$; other points are $(5,5),(7,-3),(11,5)$.
For $f(x)=x^{2}-6x + 4$, x - intercepts are $(3+\sqrt{5},0)$ and $(3 - \sqrt{5},0)$; other points are $(3,-5),(1,-1),(6,4)$.