y=-15x2+25x-7
To begin, you have provided a quadratic equation in the form y = -15x^2 + 25x - 7.
Step 1: Identify the quadratic equation's coefficients.
The coefficient of x^2 in this equation is -15, the coefficient of x is 25, and the constant term is -7.
Step 2: Determine the vertex of the quadratic equation.
To find the vertex of the quadratic equation, you can use the formula x = -b / (2a). In this case, a = -15 and b = 25.
x = -25 / (2 * -15)
x = -25 / -30
x = 5 / 6
Now substitute this value of x back into the original equation to find the y-coordinate of the vertex:
y = -15(5/6)^2 + 25(5/6) - 7
y = -15(25/36) + 125/6 - 7
y = -25/12 + 125/6 - 7
y = -25/12 + 250/12 - 7
y = (250 - 25 - 84) / 12
y = 141/12
y = 47/4
Therefore, the vertex of the quadratic equation y = -15x^2 + 25x - 7 is (5/6, 47/4).
Step 3: Determine the axis of symmetry.
The axis of symmetry is a vertical line passing through the vertex. Using the x-coordinate of the vertex, the axis of symmetry equation is x = 5/6.
Step 4: Find the y-intercept.
The y-intercept is the value of y when x is equal to 0. Substituting x = 0 into the equation, we have:
y = -15(0)^2 + 25(0) - 7
y = 0 - 0 - 7
y = -7
Hence, the y-intercept is -7.
Step 5: Determine the x-intercepts (if any).
To find the x-intercepts, set y equal to zero and solve for x.
0 = -15x^2 + 25x - 7
You can either factor the quadratic equation or use the quadratic formula to solve for x.
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation, a = -15, b = 25, and c = -7, we get:
x = (-25 ± √(25^2 - 4(-15)(-7))) / (2 * -15 )
x = (-25 ± √(625 - 420)) / -30
x = (-25 ± √205) / -30
Therefore, the x-intercepts are at (approximately) x = 2.38 and x = -0.38.
To summarize:
- The vertex of the quadratic equation y = -15x^2 + 25x - 7 is (5/6, 47/4).
- The equation of the axis of symmetry is x = 5/6.
- The y-intercept is -7.
- The x-intercepts are approximately x = 2.38 and x = -0.38.
The equation you provided is in the form of a quadratic equation, which is a polynomial equation of degree 2. The general form of a quadratic equation is given by:
ax^2 + bx + c = 0
In your equation, the coefficients are as follows:
a = -15
b = 25
c = -7
To find the solutions for the equation, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Substituting the values from your equation into the quadratic formula, we get:
x = (-25 ± √(25^2 - 4(-15)(-7)))/(2(-15))
Let's simplify this expression step by step:
First, let's calculate the discriminant, which is the expression under the square root:
b^2 - 4ac = 25^2 - 4(-15)(-7) = 625 - 420 = 205
Now, substitute the discriminant into the quadratic formula:
x = (-25 ± √205)/(2(-15))
Simplifying further:
x = (-25 ± √205)/(-30)
which can be written as:
x = (25 ± √205)/30
So, your solutions for the equation y = -15x^2 + 25x - 7 are:
x = (25 + √205)/30
x = (25 - √205)/30