Part A: x = -5/4, 3 || (-5/4, 0) (3, 0)
To find the x-intercepts, we need to know where y is equal to 0. So, we will set the function equal to 0 and solve for x.
4x^2 - 7x - 15 = 0
4 x 15 = 60 || -12 x 5 = 60 || -12 + 5 = -7
4x^2 - 12x + 5x - 15 = 0
4x(x - 3) + 5(x - 3) = 0
(4x + 5)(x - 3) = 0
4x + 5 = 0
x = -5/4
x - 3 = 0
x = 3
Part B: minimum, (7/8, -289/16)
The vertex of the graph will be a minimum. This is because the parabola is positive, meaning that it opens to the top.
To find the coordinates of the parabola, we start with the x-coordinate. The x-coordinate can be found using the equation -b/2a.
b = -7
a = 4
x = -(-7) / 2(4) = 7/8
Now that we know the x-value, we can plug it into the function and solve for the y-value.
y = 4(7/8)^2 - 7(7/8) - 15
y = 4(49/64) - 49/8 - 15
y = 196/64 - 392/64 - 960/64
y = -1156/64 = -289/16 = -18 1/16
Part C:
First, start by graphing the vertex. Then, use the x-intercepts and graph those. At this point we should have three points in a sort of triangle shape. If we did it right, each of the x-values will be an equal distance from the vertex. After we have those points graphed, it is time to draw in the parabola. Knowing that the parabola is positive, we draw in a U shape that passes through each of the three points and opens toward the top of the coordinate grid.
Hope this helps!
Answer:
The area of the triangle is 18 square units.
Step-by-step explanation:
First, we determine the lengths of segments AB, BC and AC by Pythagorean Theorem:
AB
![AB = \sqrt{(5-2)^{2}+[6-(-1)]^{2}}](https://tex.z-dn.net/?f=AB%20%3D%20%5Csqrt%7B%285-2%29%5E%7B2%7D%2B%5B6-%28-1%29%5D%5E%7B2%7D%7D)
BC
AC
![AC = \sqrt{(-1-2)^{2}+[4-(-1)]^{2}}](https://tex.z-dn.net/?f=AC%20%3D%20%5Csqrt%7B%28-1-2%29%5E%7B2%7D%2B%5B4-%28-1%29%5D%5E%7B2%7D%7D)
Now we determine the area of the triangle by Heron's formula:
(1)
(2)
Where:
- Area of the triangle.
- Semiparameter.
If we know that , and , then the area of the triangle is:
The area of the triangle is 18 square units.
Answer:
x = 1, y = 10
Step-by-step explanation:
y = -5x + 15 --- Equation 1
2x + y = 12 --- Equation 2
Substitute y = -5x + 15 into Equation 2:
2x + y = 12
2x - 5x + 15 = 12
Evaluate like terms.
15 - 3x = 12
Isolate -3x.
-3x = 12 - 15
Evaluate like terms.
-3x = -3
Find x.
x = -3 ÷ -3
x = 1
Substitute x = 1 into Equation 2:
2x + y = 12
2(1) + y = 12
2 + y = 12
Isolate y.
y = 12 - 2
y = 10
Where's the question 12 a =
