The equation of the tangent to y = -2x^2 + 3 at x : 1 is

To find the equation of the tangent to the curve \( y = -2x^2 + 3 \) at the point where \( x = 1 \), we follow these steps:

### Step 1: Find the derivative of the function
The derivative \( \frac{dy}{dx} \) gives the slope of the tangent line at any point on the curve.

Given:
\[
y = -2x^2 + 3
\]
Differentiate \( y \) with respect to \( x \):
\[
\frac{dy}{dx} = \frac{d}{dx} (-2x^2 + 3) = -4x
\]

### Step 2: Find the slope of the tangent at \( x = 1 \)
Substitute \( x = 1 \) into the derivative to get the slope at that point:
\[
\text{slope} = -4(1) = -4
\]

### Step 3: Find the coordinates of the point on the curve at \( x = 1 \)
Substitute \( x = 1 \) into the original equation to get the corresponding \( y \)-coordinate:
\[
y = -2(1)^2 + 3 = -2 + 3 = 1
\]
So the point of tangency is \( (1, 1) \).

### Step 4: Use the point-slope form of the equation of a line
The equation of a line with slope \( m \) passing through point \( (x_1, y_1) \) is given by:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( m = -4 \), \( x_1 = 1 \), and \( y_1 = 1 \):
\[
y - 1 = -4(x - 1)
\]

### Step 5: Simplify the equation
\[
y - 1 = -4x + 4
\]
\[
y = -4x + 5
\]

Thus, the equation of the tangent to the curve at \( x = 1 \) is:
\[
y = -4x + 5
\]

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Would you like a detailed explanation of any step or have questions about this process?

Here are 5 related questions you might find useful:
1. How do you find the tangent to a curve at any given point?
2. What is the geometric interpretation of a derivative?
3. How do you compute the slope of a tangent line from a function?
4. Can there be multiple tangents to a parabola at a single point?
5. How does the second derivative help in determining concavity near a tangent?

**Tip:** Always double-check if the point of tangency lies on the original curve by substituting the \( x \)-coordinate into the curve equation.

Find the equation of the tangent to the quadratic function y = -2x^2 + 3 at x = 1 using calculus. This step-by-step guide includes finding the derivative, calculating the slope at the given point, and applying the point-slope form. Suitable for students in Grades 10-12.

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