find rules for quadratic functions with graphs meeting these conditions.
1.x-intercepts at (4,0)and(-1,0) and opening upward
2.x-intercepts at (2,0)and(6,0) and maximum point=(4,12)
3.only one x-intercept and that=(-3,0) and the y-intercept=(0,18)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
in general, a quadratic equation is in the form
ax^2 + bx + c = 0.
So from the conditions given, you have to determine the value of a, b and c.
1. the function has x-intercepts at (4,0)and(-1,0), so we can simply write the function as f(x)=a(x-4)(x+1). also, it is mentioned that the function has an opening upward, so a is required to be a positive value
2.in the same way of number 1, we can write this function as f(x)=a(x-2)(x-6). notice that (4,12) is on the graph of the function, and therefore f(x) satisfies that a(4-2)(4-6)=12, so a=-3. consequently, the function is f(x)=-3(x-2)(x-6)=-3x^2+24x-36
Not the answer you are looking for? Search for more explanations.
1.opening upward, means a>0. It crosses two points (4,0) and (-1,0), so we can write y = a(x-4)(x+1)
2. y = a(x-2)(x-6) because it passes the points (2,0) and (6,0). Since it has maximum, then a<0 and since the maximum happens at (4,12), we can determine a by solving a(4-2)(4-6) = 12.
3. The quadratic graph only touch the x-intercept at (-3,0), so it has the form y = a(x+3)^2. Then it also passes through the point (0,18), we can determine a by solving 18 = a(0+3)^2
another solution for 3: why not set the function as f(x)=ax^2+bx+18 because of the y-intercept. now we have to imagine the that (-3,0) should be the lowest point of the graph of the function. so -b/2a=-3, (4a*18-b^2)/4a=0. a little bit more difficult, but ok as well. a=2, b=12. f(x)=2x^2+12x+18